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I have some problems in finding the values of series that follow this pattern: $$\sum \limits_{n=0}^{\infty} (-1)^{n}*..$$

For example: I have to find the value of this series $$\sum \limits_{n=1}^{\infty} (-1)^{n}\frac{n^2}{2^n}$$ Can you give me some tips on how I should calculate the value of this kind of series? Thank you.

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    $\begingroup$ This type of sums are called polylogarithms. $\endgroup$
    – Lucian
    Commented May 23, 2014 at 10:00
  • $\begingroup$ Well congratulations. Within $15$ minutes you received $4$ answers. As if they were all waiting for this :) $\endgroup$
    – drhab
    Commented May 23, 2014 at 10:05
  • $\begingroup$ It's amusing the users spend a great effort to get the answers but at the end of them we don't see clearly that the answer is $\displaystyle{\large -\,{2 \over 27}}$. I see this pattern frequently in $\tt M.SE$. $\endgroup$ Commented Jul 8, 2014 at 1:05

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$$\sum_{n=1}^\infty(-1)^n\frac{n^2}{2^n}=\sum_{n=1}^\infty n^2\left(-\frac12\right)^n$$

Now, $$\sum_{n=0}^\infty r^n=\frac1{1-r}$$ for $|r|<1$

Differentiate either sides to get $$\sum_{n=1}^\infty nr^{n-1}=\frac1{(1-r)^2}$$

Multiplying by $r,$ $$\implies\sum_{n=1}^\infty nr^n=\frac r{(1-r)^2}$$

Again Differentiate either sides and multiply by $r$

Set $r=-\dfrac12$

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  • $\begingroup$ typo: start second formula (in $r$) with $n=0$ instead of $n=1$ (I made the same mistake) $\endgroup$
    – drhab
    Commented May 23, 2014 at 10:00
  • $\begingroup$ @drhab, We can safely ignore $n=0,$ right? $\endgroup$ Commented May 23, 2014 at 10:06
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    $\begingroup$ Well, not in formulas like $\frac{1}{1-r}=\sum_{n=1}^{\infty}r^{n}$. It is simply not true. $\endgroup$
    – drhab
    Commented May 23, 2014 at 10:10
  • $\begingroup$ @drhab, Thanks, rectified $\endgroup$ Commented May 23, 2014 at 16:49
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Hint: start with $$f(x)=\sum_{n=0}^\infty(-x)^n=\cdots$$ and calculate $$f'(x)=\sum_{n=0}^\infty n(-x)^{n-1}(-1)=\cdots$$ $$f''(x)=\cdots$$

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Hints:

We recognize $\sum_{n=1}^{\infty}n^{2}z^{n}$ where $z=-\frac{1}{2}$.

Wellknown is $\left(1-z\right)^{-1}=\sum_{n=0}^{\infty}z^{n}$ under condition $\left|z\right|<1$.

What happens if you differentiate (twice) on both sides?

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Let $f(x):=\sum_{n\ge 1}n^2x^n$ denotes the formal power series for $n^2$. Then you need to evaluate $f(-1/2).$ Now, let $g(x)=\sum_{n\ge 1}nx^n$. Then, $f(x)-g(x)=\sum_{n\ge 1}n(n-1)x^n=x^2\sum_{n\ge 2}n(n-1)x^{n-2}$Now, we know the formal power series$$\sum_{n\ge 0}x^n=\frac{1}{1-x}\\\Rightarrow \sum_{n\ge 1}nx^{n-1}=\frac{1}{(1-x)^2} \\ \Rightarrow \sum_{n\ge 2}n(n-1)x^{n-2}=\frac{2}{(1-x)^3}$$ So, $$g(x)=\frac{x}{(1-x)^2}\\ f(x)-g(x)=\frac{2x^2}{(1-x)^3}\\ \Rightarrow f(x)=\frac{x+x^2}{(1-x)^3}$$

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