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I need to calculate the following formula

$$sum_{n=1}^{infty} 2frac{3^{n-1}n(n-1)}{(n+2)(n+1)!}$$

I know that this sum should be finite, but I don't know how to calculate it. I've done some simplifications and reductions but it get me nowhere. Please help me to calculate it.

You may proceed as follows:

• $e^x = sum_{n=0}^{infty}frac{x^n}{n!} Rightarrow e^x -(1+x+frac{x^2}{2}) = sum_{color{blue}{n=3}}^{infty} frac{x^n}{n!} = sum_{color{blue}{n=1}}^{infty} frac{x^{n+2}}{(n+2)!}$


• $Rightarrow f(x) = frac{e^x -(1+x+frac{x^2}{2})}{x^2} = sum_{n=1}^{infty} frac{x^{n}}{(n+2)!}$


• 
  $Rightarrow 2xf''(x) = boxed{2sum_{n=1}^{infty} frac{x^{n-1}n(n-1)}{(n+2)!}}$. This is your sum for $x=3$.


• $Rightarrow 2xf''(x) = 2x frac{e^x(x^2-4x+6)-2(x+3)}{x^4}= 2frac{e^x(x^2-4x+6)-2(x+3)}{x^3}$


• $x= 3 Rightarrow boxed{2sum_{n=1}^{infty} frac{3^{n-1}n(n-1)}{(n+2)!} =frac{2}{9}(e^3-4)}$

Hint:

Write $n(n-1)=(n+2)(n+1)+A(n+2)+B$

$iff n^2-n=n^2+n(3+A)+2A+B+2$

$3+A=-1iff A=-4,2A+B+2=0iff B=-2-2A=-2-2(-4)=6$

Use $sum_{r=0}^inftydfrac{y^r}{r!}=e^y$

$$dfrac{x^{n-1}n(n-1)}{(n+2)!}$$

$$=x^{n-1}cdotdfrac{ (n+2)(n+1)-4(n+2)+6}{(n+2)!} =dfrac1xcdotdfrac{x^n}{n!}-dfrac4{x^2}cdotdfrac{x^{n+1}}{(n+1)!}+dfrac6{x^3}cdotdfrac{x^{n+2}}{(n+2)!}$$

$$impliessum_{n=1}^inftydfrac{x^{n-1}n(n-1)}{(n+2)!}=dfrac1xsum_{n=1}^inftydfrac{x^n}{n!}-dfrac4{x^2}sum_{n=1}^inftydfrac{x^{n+1}}{(n+1)!}+dfrac6{x^3}sum_{n=1}^inftydfrac{x^{n+2}}{(n+2)!}$$

$$=dfrac1xleft(e^x-1right)-dfrac4{x^2}left(e^x-1-dfrac x1right)+dfrac6{x^3}left(e^x-1-dfrac x1-dfrac{x^2}{2!}right)$$

$$=e^xleft(dfrac1x-dfrac4{x^2}+dfrac6{x^3}right)+dfrac1xleft(-1+4-3right)+dfrac1{x^2}left(4+6right)-dfrac6{x^3}$$

Here $x=3$