Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Evaluate:
$$\sum\limits_{n=1}^\infty \frac{n^2}{3^n}.$$

By the ratio test, $\displaystyle\lim_{n\to\infty} \frac{(n+1)^2}{3^{n+1}}\cdot\frac{3^n}{n^2}=1/3,$ which is less than 1, therefore the series is convergent.

Now I am stuck on how to evaluate this series, without the $n^2$ on top, it can be easily calculated by the geometric series formula. Any help would be appreciated.

share|improve this question
4  
See here for a solution technique; you'll have to differentiate twice, since you have $n^2$. –  user61527 Mar 15 at 19:25
5  
Write down the series for $1/(1-x)$. Differentiate. Multiply by $x$. Differentiate again. –  David Mitra Mar 15 at 19:28
    
Thank you two so much! –  user73645 Mar 15 at 19:55
    
See polylogarithm. –  Lucian Mar 15 at 20:38
    

1 Answer 1

up vote 3 down vote accepted

Just to expand on David's comment: \begin{align*} \frac{1}{1-x} &= 1 + x + x^2 + \cdots + x^n + \cdots \\ \frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{1}{(1-x)^2} &= 1 + 2x + 3x^2 + \cdots + nx^{n-1} + \cdots \\ \frac{x}{(1-x)^2} &= x + 2x^2 + 3x^3 + \cdots + nx^{n} + \cdots \\ \frac{d}{dx} \left( \frac{x}{(1-x)^2} \right) = \frac{1 + x}{(1 - x)^3} &= 1 + 2^2x + 3^2x^2 + \cdots + n^2x^{n-1} + \cdots \\ \frac{x(1 + x)}{(1 - x)^3} &= x + 2^2x^2 + 3^2x^3 + \cdots + n^2x^{n} + \cdots \\ \end{align*} Now plug in $x = \frac13$.

share|improve this answer
    
Thank you so much! I didn't know there is such a trick! –  user73645 Mar 15 at 19:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.