Is there a “greatest function” that converges?












18












$begingroup$


We just hit convergence tests in calculus, and learned that $sum_{n=1}^{infty} frac{1}{n^p}$
converges for all $p gt 1$. I thought that this was sort of a "barrier" between what converges and what diverges. Specifically, that setting $a_n=frac{1}{n^{1+epsilon}}$ is sort of the "greatest function" (I'll make this precise later) for which $sum a_n$ converges.



But, I did realize that there are functions that dominate $frac{1}{n^{1+epsilon}}$ but not $frac1n$, such as $frac{1}{nlog(n)}$. Now, the sum of that specific example diverges, but it got me wondering about whether $frac{1}{n}$ is truly the "boundary". So, this leads me to two questions.



1) Is there a function $f$ that dominates $frac{1}{n^p}$ for all $p>1$, meaning: $$lim_{xtoinfty} frac{f(x)}{frac{1}{x^p}}=infty$$
Such that:
$$sum_{n=1}^infty f(n)$$ converges?



2) If so, up to a constant is there a function $g$ such that $sum_{n=1}^infty g(n)$ converges, such that $g$ dominates $f$ for all other functions $f$ such that $sum_{n=1}^infty f(n)$ converges?



I'm just a freshman in high school so I apologize if this is a stupid question.






calculus sequences-and-series convergence







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
    $endgroup$
    – Electric Moccasins
    9 hours ago












  • $begingroup$
    How about $f(x) = frac{1}{xln^2(x)}$ for your first question
    $endgroup$
    – Jakobian
    9 hours ago








  • 5




    $begingroup$
    It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
    $endgroup$
    – Henning Makholm
    9 hours ago






  • 1




    $begingroup$
    From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
    $endgroup$
    – Martin R
    9 hours ago
















18












$begingroup$


We just hit convergence tests in calculus, and learned that $sum_{n=1}^{infty} frac{1}{n^p}$
converges for all $p gt 1$. I thought that this was sort of a "barrier" between what converges and what diverges. Specifically, that setting $a_n=frac{1}{n^{1+epsilon}}$ is sort of the "greatest function" (I'll make this precise later) for which $sum a_n$ converges.



But, I did realize that there are functions that dominate $frac{1}{n^{1+epsilon}}$ but not $frac1n$, such as $frac{1}{nlog(n)}$. Now, the sum of that specific example diverges, but it got me wondering about whether $frac{1}{n}$ is truly the "boundary". So, this leads me to two questions.



1) Is there a function $f$ that dominates $frac{1}{n^p}$ for all $p>1$, meaning: $$lim_{xtoinfty} frac{f(x)}{frac{1}{x^p}}=infty$$
Such that:
$$sum_{n=1}^infty f(n)$$ converges?



2) If so, up to a constant is there a function $g$ such that $sum_{n=1}^infty g(n)$ converges, such that $g$ dominates $f$ for all other functions $f$ such that $sum_{n=1}^infty f(n)$ converges?



I'm just a freshman in high school so I apologize if this is a stupid question.






calculus sequences-and-series convergence







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
    $endgroup$
    – Electric Moccasins
    9 hours ago












  • $begingroup$
    How about $f(x) = frac{1}{xln^2(x)}$ for your first question
    $endgroup$
    – Jakobian
    9 hours ago








  • 5




    $begingroup$
    It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
    $endgroup$
    – Henning Makholm
    9 hours ago






  • 1




    $begingroup$
    From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
    $endgroup$
    – Martin R
    9 hours ago














18












18








18


1



$begingroup$


We just hit convergence tests in calculus, and learned that $sum_{n=1}^{infty} frac{1}{n^p}$
converges for all $p gt 1$. I thought that this was sort of a "barrier" between what converges and what diverges. Specifically, that setting $a_n=frac{1}{n^{1+epsilon}}$ is sort of the "greatest function" (I'll make this precise later) for which $sum a_n$ converges.



But, I did realize that there are functions that dominate $frac{1}{n^{1+epsilon}}$ but not $frac1n$, such as $frac{1}{nlog(n)}$. Now, the sum of that specific example diverges, but it got me wondering about whether $frac{1}{n}$ is truly the "boundary". So, this leads me to two questions.



1) Is there a function $f$ that dominates $frac{1}{n^p}$ for all $p>1$, meaning: $$lim_{xtoinfty} frac{f(x)}{frac{1}{x^p}}=infty$$
Such that:
$$sum_{n=1}^infty f(n)$$ converges?



2) If so, up to a constant is there a function $g$ such that $sum_{n=1}^infty g(n)$ converges, such that $g$ dominates $f$ for all other functions $f$ such that $sum_{n=1}^infty f(n)$ converges?



I'm just a freshman in high school so I apologize if this is a stupid question.






calculus sequences-and-series convergence







share|cite|improve this question









$endgroup$




We just hit convergence tests in calculus, and learned that $sum_{n=1}^{infty} frac{1}{n^p}$
converges for all $p gt 1$. I thought that this was sort of a "barrier" between what converges and what diverges. Specifically, that setting $a_n=frac{1}{n^{1+epsilon}}$ is sort of the "greatest function" (I'll make this precise later) for which $sum a_n$ converges.



But, I did realize that there are functions that dominate $frac{1}{n^{1+epsilon}}$ but not $frac1n$, such as $frac{1}{nlog(n)}$. Now, the sum of that specific example diverges, but it got me wondering about whether $frac{1}{n}$ is truly the "boundary". So, this leads me to two questions.



1) Is there a function $f$ that dominates $frac{1}{n^p}$ for all $p>1$, meaning: $$lim_{xtoinfty} frac{f(x)}{frac{1}{x^p}}=infty$$
Such that:
$$sum_{n=1}^infty f(n)$$ converges?



2) If so, up to a constant is there a function $g$ such that $sum_{n=1}^infty g(n)$ converges, such that $g$ dominates $f$ for all other functions $f$ such that $sum_{n=1}^infty f(n)$ converges?



I'm just a freshman in high school so I apologize if this is a stupid question.






calculus sequences-and-series convergence




calculus sequences-and-series convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 9 hours ago









Electric MoccasinsElectric Moccasins

1163




1163








  • 1




    $begingroup$
    I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
    $endgroup$
    – Electric Moccasins
    9 hours ago












  • $begingroup$
    How about $f(x) = frac{1}{xln^2(x)}$ for your first question
    $endgroup$
    – Jakobian
    9 hours ago








  • 5




    $begingroup$
    It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
    $endgroup$
    – Henning Makholm
    9 hours ago






  • 1




    $begingroup$
    From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
    $endgroup$
    – Martin R
    9 hours ago














  • 1




    $begingroup$
    I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
    $endgroup$
    – Electric Moccasins
    9 hours ago












  • $begingroup$
    How about $f(x) = frac{1}{xln^2(x)}$ for your first question
    $endgroup$
    – Jakobian
    9 hours ago








  • 5




    $begingroup$
    It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
    $endgroup$
    – Henning Makholm
    9 hours ago






  • 1




    $begingroup$
    From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
    $endgroup$
    – Martin R
    9 hours ago








1




1




$begingroup$
I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
$endgroup$
– Electric Moccasins
9 hours ago






$begingroup$
I tried to format the question such that it excludes all the "trivial" answers, i.e. "no, because you can just take this function and multiply by any constant to get another function that dominates it and still converges". The constant thing is just excluded by the fact that the limit has to diverge, rather than be equal to some constant, but something similar could still possibly happen. I tried to include a sort of "parameter" on $f$ to patch this, but I couldn't get it formal. But the sort of "philosophy" of the question is about the form of the function, rather than a "cop-out" like that.
$endgroup$
– Electric Moccasins
9 hours ago














$begingroup$
How about $f(x) = frac{1}{xln^2(x)}$ for your first question
$endgroup$
– Jakobian
9 hours ago






$begingroup$
How about $f(x) = frac{1}{xln^2(x)}$ for your first question
$endgroup$
– Jakobian
9 hours ago






5




5




$begingroup$
It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
$endgroup$
– Henning Makholm
9 hours ago




$begingroup$
It's a good question; part of a good answer will surely be to identify the most fruitful way to make the vague intuition behind it more precise.
$endgroup$
– Henning Makholm
9 hours ago




1




1




$begingroup$
From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
$endgroup$
– Martin R
9 hours ago




$begingroup$
From math.stackexchange.com/a/452074/42969: Let $sum_{n=1}^{infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $sum_{n=1}^{infty} C_n$ with much bigger terms in the sense that $lim_{nrightarrowinfty} C_n/c_n = infty$.
$endgroup$
– Martin R
9 hours ago










2 Answers
2






active

oldest

votes


















8












$begingroup$

1) Yes, $f(n)=frac{1}{n(ln{n})^2}$.



2) No. Assume $f geq 0$ and $sum_{n geq 1}{f(n)} < infty$.



Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $sum_{N_n+1}^{N_{n+1}}{f(k)} leq C2^{-n}$.



Then set $g(n)=(p+1)f(n)$, where $N_p < n leq N_{p+1}$.



Then $sum_{N_n+1}^{N_{n+1}}{g(k)} leq C(n+1)2^{-n}$, thus $sum_{n geq 1}{g(n)}$ is finite and $g(n) >> f(n)$.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    1) $$f(n) = frac{1}{n log(n)^2}$$



    2) No. Given any $g > 0$ such that $sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$sum_{n ge M_k} g(n) < 2^{-k}$$

    Then
    $ sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k le n < M_{k+1}$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
      $endgroup$
      – orlp
      9 hours ago












    • $begingroup$
      Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
      $endgroup$
      – Electric Moccasins
      6 hours ago










    • $begingroup$
      @ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
      $endgroup$
      – Solomonoff's Secret
      5 hours ago












    • $begingroup$
      @Solomonoff'sSecret Ok, I'm really not sure how you're getting that inequality. I get the inequality in Prof. Isreal's answer, but not how it translates to the one you said. Sorry, this is my first ever brush with "hard math".
      $endgroup$
      – Electric Moccasins
      4 hours ago












    • $begingroup$
      $$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
      $endgroup$
      – Robert Israel
      1 hour ago













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8












    $begingroup$

    1) Yes, $f(n)=frac{1}{n(ln{n})^2}$.



    2) No. Assume $f geq 0$ and $sum_{n geq 1}{f(n)} < infty$.



    Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $sum_{N_n+1}^{N_{n+1}}{f(k)} leq C2^{-n}$.



    Then set $g(n)=(p+1)f(n)$, where $N_p < n leq N_{p+1}$.



    Then $sum_{N_n+1}^{N_{n+1}}{g(k)} leq C(n+1)2^{-n}$, thus $sum_{n geq 1}{g(n)}$ is finite and $g(n) >> f(n)$.






    share|cite|improve this answer











    $endgroup$


















      8












      $begingroup$

      1) Yes, $f(n)=frac{1}{n(ln{n})^2}$.



      2) No. Assume $f geq 0$ and $sum_{n geq 1}{f(n)} < infty$.



      Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $sum_{N_n+1}^{N_{n+1}}{f(k)} leq C2^{-n}$.



      Then set $g(n)=(p+1)f(n)$, where $N_p < n leq N_{p+1}$.



      Then $sum_{N_n+1}^{N_{n+1}}{g(k)} leq C(n+1)2^{-n}$, thus $sum_{n geq 1}{g(n)}$ is finite and $g(n) >> f(n)$.






      share|cite|improve this answer











      $endgroup$
















        8












        8








        8





        $begingroup$

        1) Yes, $f(n)=frac{1}{n(ln{n})^2}$.



        2) No. Assume $f geq 0$ and $sum_{n geq 1}{f(n)} < infty$.



        Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $sum_{N_n+1}^{N_{n+1}}{f(k)} leq C2^{-n}$.



        Then set $g(n)=(p+1)f(n)$, where $N_p < n leq N_{p+1}$.



        Then $sum_{N_n+1}^{N_{n+1}}{g(k)} leq C(n+1)2^{-n}$, thus $sum_{n geq 1}{g(n)}$ is finite and $g(n) >> f(n)$.






        share|cite|improve this answer











        $endgroup$



        1) Yes, $f(n)=frac{1}{n(ln{n})^2}$.



        2) No. Assume $f geq 0$ and $sum_{n geq 1}{f(n)} < infty$.



        Then there exists an increasing sequence $N_n$ and some constant $C > 0$ such that $sum_{N_n+1}^{N_{n+1}}{f(k)} leq C2^{-n}$.



        Then set $g(n)=(p+1)f(n)$, where $N_p < n leq N_{p+1}$.



        Then $sum_{N_n+1}^{N_{n+1}}{g(k)} leq C(n+1)2^{-n}$, thus $sum_{n geq 1}{g(n)}$ is finite and $g(n) >> f(n)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 5 hours ago









        user12345

        2,0231515




        2,0231515










        answered 9 hours ago









        MindlackMindlack

        3,62517




        3,62517























            3












            $begingroup$

            1) $$f(n) = frac{1}{n log(n)^2}$$



            2) No. Given any $g > 0$ such that $sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$sum_{n ge M_k} g(n) < 2^{-k}$$

            Then
            $ sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k le n < M_{k+1}$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
              $endgroup$
              – orlp
              9 hours ago












            • $begingroup$
              Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
              $endgroup$
              – Electric Moccasins
              6 hours ago










            • $begingroup$
              @ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
              $endgroup$
              – Solomonoff's Secret
              5 hours ago












            • $begingroup$
              @Solomonoff'sSecret Ok, I'm really not sure how you're getting that inequality. I get the inequality in Prof. Isreal's answer, but not how it translates to the one you said. Sorry, this is my first ever brush with "hard math".
              $endgroup$
              – Electric Moccasins
              4 hours ago












            • $begingroup$
              $$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
              $endgroup$
              – Robert Israel
              1 hour ago


















            3












            $begingroup$

            1) $$f(n) = frac{1}{n log(n)^2}$$



            2) No. Given any $g > 0$ such that $sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$sum_{n ge M_k} g(n) < 2^{-k}$$

            Then
            $ sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k le n < M_{k+1}$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
              $endgroup$
              – orlp
              9 hours ago












            • $begingroup$
              Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
              $endgroup$
              – Electric Moccasins
              6 hours ago










            • $begingroup$
              @ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
              $endgroup$
              – Solomonoff's Secret
              5 hours ago












            • $begingroup$
              @Solomonoff'sSecret Ok, I'm really not sure how you're getting that inequality. I get the inequality in Prof. Isreal's answer, but not how it translates to the one you said. Sorry, this is my first ever brush with "hard math".
              $endgroup$
              – Electric Moccasins
              4 hours ago












            • $begingroup$
              $$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
              $endgroup$
              – Robert Israel
              1 hour ago
















            3












            3








            3





            $begingroup$

            1) $$f(n) = frac{1}{n log(n)^2}$$



            2) No. Given any $g > 0$ such that $sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$sum_{n ge M_k} g(n) < 2^{-k}$$

            Then
            $ sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k le n < M_{k+1}$.






            share|cite|improve this answer









            $endgroup$



            1) $$f(n) = frac{1}{n log(n)^2}$$



            2) No. Given any $g > 0$ such that $sum_n g(n)$ converges, there is an increasing sequence $M_k$ such that $$sum_{n ge M_k} g(n) < 2^{-k}$$

            Then
            $ sum_n g(n) h(n)$ converges, where $h(n) = k$ for $M_k le n < M_{k+1}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 9 hours ago









            Robert IsraelRobert Israel

            322k23212465




            322k23212465








            • 1




              $begingroup$
              If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
              $endgroup$
              – orlp
              9 hours ago












            • $begingroup$
              Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
              $endgroup$
              – Electric Moccasins
              6 hours ago










            • $begingroup$
              @ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
              $endgroup$
              – Solomonoff's Secret
              5 hours ago












            • $begingroup$
              @Solomonoff'sSecret Ok, I'm really not sure how you're getting that inequality. I get the inequality in Prof. Isreal's answer, but not how it translates to the one you said. Sorry, this is my first ever brush with "hard math".
              $endgroup$
              – Electric Moccasins
              4 hours ago












            • $begingroup$
              $$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
              $endgroup$
              – Robert Israel
              1 hour ago
















            • 1




              $begingroup$
              If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
              $endgroup$
              – orlp
              9 hours ago












            • $begingroup$
              Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
              $endgroup$
              – Electric Moccasins
              6 hours ago










            • $begingroup$
              @ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
              $endgroup$
              – Solomonoff's Secret
              5 hours ago












            • $begingroup$
              @Solomonoff'sSecret Ok, I'm really not sure how you're getting that inequality. I get the inequality in Prof. Isreal's answer, but not how it translates to the one you said. Sorry, this is my first ever brush with "hard math".
              $endgroup$
              – Electric Moccasins
              4 hours ago












            • $begingroup$
              $$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
              $endgroup$
              – Robert Israel
              1 hour ago










            1




            1




            $begingroup$
            If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
            $endgroup$
            – orlp
            9 hours ago






            $begingroup$
            If we're being pedantic you need $f(n) = dfrac{1}{nlog(n+1)^2}$ (or something similar) to prevent division by zero.
            $endgroup$
            – orlp
            9 hours ago














            $begingroup$
            Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
            $endgroup$
            – Electric Moccasins
            6 hours ago




            $begingroup$
            Can you expand on how you know $sum g(n)h(n)$ converges? Just trying to wrap my head around this
            $endgroup$
            – Electric Moccasins
            6 hours ago












            $begingroup$
            @ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
            $endgroup$
            – Solomonoff's Secret
            5 hours ago






            $begingroup$
            @ElectricMoccasins Because $sum g(n) h(n) le sum k 2^{-k}$, which converges.
            $endgroup$
            – Solomonoff's Secret
            5 hours ago














            $begingroup$
            @Solomonoff'sSecret Ok, I'm really not sure how you're getting that inequality. I get the inequality in Prof. Isreal's answer, but not how it translates to the one you said. Sorry, this is my first ever brush with "hard math".
            $endgroup$
            – Electric Moccasins
            4 hours ago






            $begingroup$
            @Solomonoff'sSecret Ok, I'm really not sure how you're getting that inequality. I get the inequality in Prof. Isreal's answer, but not how it translates to the one you said. Sorry, this is my first ever brush with "hard math".
            $endgroup$
            – Electric Moccasins
            4 hours ago














            $begingroup$
            $$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
            $endgroup$
            – Robert Israel
            1 hour ago






            $begingroup$
            $$sum_{n=M_k}^{M_{k+1}-1} g(n) h(n) = k sum_{n=M_k}^{M_{k+1}-1} g(n) < k 2^{-k}$$
            $endgroup$
            – Robert Israel
            1 hour ago




















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