Showing a sum is positive
$begingroup$
Show that the sum$$sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}$$ is a positive rational number.
It is easy to show that it is a rational number. But I am having trouble showing that this expression is positive. It might be some binomial expansion that I could not get.
combinatorics summation binomial-coefficients binomial-ideals
$endgroup$
add a comment |
$begingroup$
Show that the sum$$sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}$$ is a positive rational number.
It is easy to show that it is a rational number. But I am having trouble showing that this expression is positive. It might be some binomial expansion that I could not get.
combinatorics summation binomial-coefficients binomial-ideals
$endgroup$
1
$begingroup$
Have you tried using induction on $n$ for example?
$endgroup$
– Minus One-Twelfth
1 hour ago
add a comment |
$begingroup$
Show that the sum$$sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}$$ is a positive rational number.
It is easy to show that it is a rational number. But I am having trouble showing that this expression is positive. It might be some binomial expansion that I could not get.
combinatorics summation binomial-coefficients binomial-ideals
$endgroup$
Show that the sum$$sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}$$ is a positive rational number.
It is easy to show that it is a rational number. But I am having trouble showing that this expression is positive. It might be some binomial expansion that I could not get.
combinatorics summation binomial-coefficients binomial-ideals
combinatorics summation binomial-coefficients binomial-ideals
edited 1 hour ago
Hitendra Kumar
asked 1 hour ago
Hitendra KumarHitendra Kumar
656
656
1
$begingroup$
Have you tried using induction on $n$ for example?
$endgroup$
– Minus One-Twelfth
1 hour ago
add a comment |
1
$begingroup$
Have you tried using induction on $n$ for example?
$endgroup$
– Minus One-Twelfth
1 hour ago
1
1
$begingroup$
Have you tried using induction on $n$ for example?
$endgroup$
– Minus One-Twelfth
1 hour ago
$begingroup$
Have you tried using induction on $n$ for example?
$endgroup$
– Minus One-Twelfth
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Direct proof:
$$begin{split}
sum_{k=0}^{n} {nchoose k}frac{{(-1)}^k}{n+k+1} &=sum_{k=0}^{n} {nchoose k}(-1)^kint_0^1 x^{n+k}dx\
&=int_0^1x^nsum_{k=0}^{n} {nchoose k}(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
end{split}$$
The latter is clearly a positive number.
$endgroup$
$begingroup$
Thanks,I got it.
$endgroup$
– Hitendra Kumar
1 hour ago
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
1 hour ago
$begingroup$
How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
$endgroup$
– NoChance
59 mins ago
2
$begingroup$
It's a "known trick" that $frac 1 {p+1} = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
$endgroup$
– Stefan Lafon
54 mins ago
$begingroup$
Thanks for responding.Got it.
$endgroup$
– NoChance
44 mins ago
add a comment |
$begingroup$
When $k=0$ the term is positive. When $k=1$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=0$ TERM.
When $k=2$ the term is positive. When $k=3$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=2$ TERM.
.....
Get it?
$endgroup$
$begingroup$
sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
$endgroup$
– Hitendra Kumar
1 hour ago
add a comment |
$begingroup$
We can specifically prove that
$$
boxed{sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}=left((2n+1)binom{2n}nright)^{-1}}
$$
To see this, shuffle a deck of $2n+1$ cards numbered $1$ to $n$. Consider this:
What is the probability that card number $n+1$ is in the middle of the deck, and cards numbered $1$ to $n$ are below it?
The easy answer is the fraction on the RHS. The LHS can be interpreted as an application of the principle of inclusion exclusion. Namely, we first take the probability that card number of $n+1$ is the lowest of the cards numbered $n+1,n+2,dots,2n+1$. This is the $k=0$ term. From this, for each $i=1,dots,n$, we subtract the probability that $n+1$ is the lowest of the list $i,n+1,n+2,dots,2n+1$. This is a bad event, because we want $n+1$ to be above $i$. Doing this for each $i$, we subtract $binom{n}1frac{1}{n+2}$. We then must add back in the doubly subtracted events, subtract the triple intersections, and so on, eventually ending with the alternating sum on the left.
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Direct proof:
$$begin{split}
sum_{k=0}^{n} {nchoose k}frac{{(-1)}^k}{n+k+1} &=sum_{k=0}^{n} {nchoose k}(-1)^kint_0^1 x^{n+k}dx\
&=int_0^1x^nsum_{k=0}^{n} {nchoose k}(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
end{split}$$
The latter is clearly a positive number.
$endgroup$
$begingroup$
Thanks,I got it.
$endgroup$
– Hitendra Kumar
1 hour ago
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
1 hour ago
$begingroup$
How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
$endgroup$
– NoChance
59 mins ago
2
$begingroup$
It's a "known trick" that $frac 1 {p+1} = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
$endgroup$
– Stefan Lafon
54 mins ago
$begingroup$
Thanks for responding.Got it.
$endgroup$
– NoChance
44 mins ago
add a comment |
$begingroup$
Direct proof:
$$begin{split}
sum_{k=0}^{n} {nchoose k}frac{{(-1)}^k}{n+k+1} &=sum_{k=0}^{n} {nchoose k}(-1)^kint_0^1 x^{n+k}dx\
&=int_0^1x^nsum_{k=0}^{n} {nchoose k}(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
end{split}$$
The latter is clearly a positive number.
$endgroup$
$begingroup$
Thanks,I got it.
$endgroup$
– Hitendra Kumar
1 hour ago
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
1 hour ago
$begingroup$
How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
$endgroup$
– NoChance
59 mins ago
2
$begingroup$
It's a "known trick" that $frac 1 {p+1} = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
$endgroup$
– Stefan Lafon
54 mins ago
$begingroup$
Thanks for responding.Got it.
$endgroup$
– NoChance
44 mins ago
add a comment |
$begingroup$
Direct proof:
$$begin{split}
sum_{k=0}^{n} {nchoose k}frac{{(-1)}^k}{n+k+1} &=sum_{k=0}^{n} {nchoose k}(-1)^kint_0^1 x^{n+k}dx\
&=int_0^1x^nsum_{k=0}^{n} {nchoose k}(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
end{split}$$
The latter is clearly a positive number.
$endgroup$
Direct proof:
$$begin{split}
sum_{k=0}^{n} {nchoose k}frac{{(-1)}^k}{n+k+1} &=sum_{k=0}^{n} {nchoose k}(-1)^kint_0^1 x^{n+k}dx\
&=int_0^1x^nsum_{k=0}^{n} {nchoose k}(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
end{split}$$
The latter is clearly a positive number.
answered 1 hour ago
Stefan LafonStefan Lafon
3,02019
3,02019
$begingroup$
Thanks,I got it.
$endgroup$
– Hitendra Kumar
1 hour ago
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
1 hour ago
$begingroup$
How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
$endgroup$
– NoChance
59 mins ago
2
$begingroup$
It's a "known trick" that $frac 1 {p+1} = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
$endgroup$
– Stefan Lafon
54 mins ago
$begingroup$
Thanks for responding.Got it.
$endgroup$
– NoChance
44 mins ago
add a comment |
$begingroup$
Thanks,I got it.
$endgroup$
– Hitendra Kumar
1 hour ago
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
1 hour ago
$begingroup$
How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
$endgroup$
– NoChance
59 mins ago
2
$begingroup$
It's a "known trick" that $frac 1 {p+1} = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
$endgroup$
– Stefan Lafon
54 mins ago
$begingroup$
Thanks for responding.Got it.
$endgroup$
– NoChance
44 mins ago
$begingroup$
Thanks,I got it.
$endgroup$
– Hitendra Kumar
1 hour ago
$begingroup$
Thanks,I got it.
$endgroup$
– Hitendra Kumar
1 hour ago
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
1 hour ago
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
1 hour ago
$begingroup$
How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
$endgroup$
– NoChance
59 mins ago
$begingroup$
How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
$endgroup$
– NoChance
59 mins ago
2
2
$begingroup$
It's a "known trick" that $frac 1 {p+1} = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
$endgroup$
– Stefan Lafon
54 mins ago
$begingroup$
It's a "known trick" that $frac 1 {p+1} = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
$endgroup$
– Stefan Lafon
54 mins ago
$begingroup$
Thanks for responding.Got it.
$endgroup$
– NoChance
44 mins ago
$begingroup$
Thanks for responding.Got it.
$endgroup$
– NoChance
44 mins ago
add a comment |
$begingroup$
When $k=0$ the term is positive. When $k=1$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=0$ TERM.
When $k=2$ the term is positive. When $k=3$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=2$ TERM.
.....
Get it?
$endgroup$
$begingroup$
sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
$endgroup$
– Hitendra Kumar
1 hour ago
add a comment |
$begingroup$
When $k=0$ the term is positive. When $k=1$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=0$ TERM.
When $k=2$ the term is positive. When $k=3$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=2$ TERM.
.....
Get it?
$endgroup$
$begingroup$
sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
$endgroup$
– Hitendra Kumar
1 hour ago
add a comment |
$begingroup$
When $k=0$ the term is positive. When $k=1$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=0$ TERM.
When $k=2$ the term is positive. When $k=3$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=2$ TERM.
.....
Get it?
$endgroup$
When $k=0$ the term is positive. When $k=1$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=0$ TERM.
When $k=2$ the term is positive. When $k=3$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=2$ TERM.
.....
Get it?
answered 1 hour ago
David G. StorkDavid G. Stork
11.1k41432
11.1k41432
$begingroup$
sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
$endgroup$
– Hitendra Kumar
1 hour ago
add a comment |
$begingroup$
sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
$endgroup$
– Hitendra Kumar
1 hour ago
$begingroup$
sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
$endgroup$
– Hitendra Kumar
1 hour ago
$begingroup$
sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
$endgroup$
– Hitendra Kumar
1 hour ago
add a comment |
$begingroup$
We can specifically prove that
$$
boxed{sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}=left((2n+1)binom{2n}nright)^{-1}}
$$
To see this, shuffle a deck of $2n+1$ cards numbered $1$ to $n$. Consider this:
What is the probability that card number $n+1$ is in the middle of the deck, and cards numbered $1$ to $n$ are below it?
The easy answer is the fraction on the RHS. The LHS can be interpreted as an application of the principle of inclusion exclusion. Namely, we first take the probability that card number of $n+1$ is the lowest of the cards numbered $n+1,n+2,dots,2n+1$. This is the $k=0$ term. From this, for each $i=1,dots,n$, we subtract the probability that $n+1$ is the lowest of the list $i,n+1,n+2,dots,2n+1$. This is a bad event, because we want $n+1$ to be above $i$. Doing this for each $i$, we subtract $binom{n}1frac{1}{n+2}$. We then must add back in the doubly subtracted events, subtract the triple intersections, and so on, eventually ending with the alternating sum on the left.
$endgroup$
add a comment |
$begingroup$
We can specifically prove that
$$
boxed{sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}=left((2n+1)binom{2n}nright)^{-1}}
$$
To see this, shuffle a deck of $2n+1$ cards numbered $1$ to $n$. Consider this:
What is the probability that card number $n+1$ is in the middle of the deck, and cards numbered $1$ to $n$ are below it?
The easy answer is the fraction on the RHS. The LHS can be interpreted as an application of the principle of inclusion exclusion. Namely, we first take the probability that card number of $n+1$ is the lowest of the cards numbered $n+1,n+2,dots,2n+1$. This is the $k=0$ term. From this, for each $i=1,dots,n$, we subtract the probability that $n+1$ is the lowest of the list $i,n+1,n+2,dots,2n+1$. This is a bad event, because we want $n+1$ to be above $i$. Doing this for each $i$, we subtract $binom{n}1frac{1}{n+2}$. We then must add back in the doubly subtracted events, subtract the triple intersections, and so on, eventually ending with the alternating sum on the left.
$endgroup$
add a comment |
$begingroup$
We can specifically prove that
$$
boxed{sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}=left((2n+1)binom{2n}nright)^{-1}}
$$
To see this, shuffle a deck of $2n+1$ cards numbered $1$ to $n$. Consider this:
What is the probability that card number $n+1$ is in the middle of the deck, and cards numbered $1$ to $n$ are below it?
The easy answer is the fraction on the RHS. The LHS can be interpreted as an application of the principle of inclusion exclusion. Namely, we first take the probability that card number of $n+1$ is the lowest of the cards numbered $n+1,n+2,dots,2n+1$. This is the $k=0$ term. From this, for each $i=1,dots,n$, we subtract the probability that $n+1$ is the lowest of the list $i,n+1,n+2,dots,2n+1$. This is a bad event, because we want $n+1$ to be above $i$. Doing this for each $i$, we subtract $binom{n}1frac{1}{n+2}$. We then must add back in the doubly subtracted events, subtract the triple intersections, and so on, eventually ending with the alternating sum on the left.
$endgroup$
We can specifically prove that
$$
boxed{sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}=left((2n+1)binom{2n}nright)^{-1}}
$$
To see this, shuffle a deck of $2n+1$ cards numbered $1$ to $n$. Consider this:
What is the probability that card number $n+1$ is in the middle of the deck, and cards numbered $1$ to $n$ are below it?
The easy answer is the fraction on the RHS. The LHS can be interpreted as an application of the principle of inclusion exclusion. Namely, we first take the probability that card number of $n+1$ is the lowest of the cards numbered $n+1,n+2,dots,2n+1$. This is the $k=0$ term. From this, for each $i=1,dots,n$, we subtract the probability that $n+1$ is the lowest of the list $i,n+1,n+2,dots,2n+1$. This is a bad event, because we want $n+1$ to be above $i$. Doing this for each $i$, we subtract $binom{n}1frac{1}{n+2}$. We then must add back in the doubly subtracted events, subtract the triple intersections, and so on, eventually ending with the alternating sum on the left.
answered 25 mins ago
Mike EarnestMike Earnest
25.5k22151
25.5k22151
add a comment |
add a comment |
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$begingroup$
Have you tried using induction on $n$ for example?
$endgroup$
– Minus One-Twelfth
1 hour ago