Sum of an Infinite Series

[jackiechad]

Here is the problem I am working on:

The sum from 1 to infinity of (n^2)/(7^n)

I believe I need to somehow arrange this into a telescoping series but have had a tough time arranging it so that the terms work out for me. I would greatly appreciate any help you can give. Thanks so much!

 

[pka]

Start with \(\displaystyle \sum\limits_{k = 1}^\infty {x^k } = \frac{x}{{1 - x}},\quad \left| x \right| < 1\).

Differenate and multiply x to get \(\displaystyle \[\sum\limits_{k = 1}^\infty {kx^k } = \frac{x}{{\left( {1 - x} \right)^2 }}\).

Do that again to get \(\displaystyle \sum\limits_{k = 1}^\infty {k^2 x^k } = \frac{{x + x^2 }}{{\left( {1 - x} \right)^3 }}\).

Then let \(\displaystyle x=\frac{1}{7}\).

 

[BigGlenntheHeavy]

\(\displaystyle Good \ show \ pka, \ I'm \ impressed.\)

 

[pka]

\(\displaystyle Good \ show \ pka, \ I'm \ impressed.\)

WHY? It is completely obvious to anyone having done similar problems.

 

[jackiechad]

Thank you SO much for the help, I was having an awful time with this problem. If you don't mind could you show the step where you differentiate, I'm a bit confused on that. Also, does plugging in 1/7 for x in the final equation give the sum? THANK YOU again!

 

[BigGlenntheHeavy]

\(\displaystyle jackiechad, \ here's \ what \ pka \ did.\)

\(\displaystyle Start \ with: \ \sum_{n=1}^{\infty}x^n \ =\frac{x}{1-x}, \ |x| \ < \ 1\)

\(\displaystyle Now, \ D_x\bigg[\sum_{n=1}^{\infty}x^n \ =\frac{x}{1-x}\bigg] \ = \ \sum_{n=1}^{\infty}nx^{n-1} \ = \ \frac{1}{(1-x)^2}\)

\(\displaystyle Then, \ x\bigg[ \sum_{n=1}^{\infty}nx^{n-1} \ = \ \frac{1}{(1-x)^2}\bigg] \ = \ \sum_{n=1}^{\infty}nx^n \ = \ \frac{x}{(1-x)^2}\)

\(\displaystyle Again, \ D_x\bigg[\sum_{n=1}^{\infty}nx^n \ = \ \frac{x}{(1-x)^2}\bigg] \ = \ \sum_{n=1}^{\infty}n^2x^{n-1} \ = \ \frac{1+x}{(1-x)^3}\)

\(\displaystyle And \ finally: \ x\bigg[\sum_{n=1}^{\infty}n^2x^{n-1} \ = \ \frac{1+x}{(1-x)^3}\bigg] \ = \ \sum_{n=1}^{\infty}n^2 x^n \ = \ \frac{x+x^2}{(1-x)^3}\)

\(\displaystyle Now, \ letting \ x \ = \ \frac{1}{7}, \ we \ get \ \sum_{n=1}^{\infty}\frac{n^2}{7^n} \ = \ \frac{\frac{1}{7}+\frac{1}{49}}{(\frac{6}{7})^3} \ = \ \frac{7}{27}.\)

 

[jackiechad]

Thank you for elaborating. This is a rather neat problem and certainly one that takes knowing that trick. Thank you both for the assistance