# Calculate the sum $\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n\times2^{2n+1}}$

I started with

$arctg(x) = \sum\limits_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}$

Then I differentiated to get rid of the denominator. Then divide with $x$ to get $x^{2n-1}$. Then integrate to get $2n$ in the denominator. Then multiply with 2 to get rid of the 2 in the denominator and finally, multiply with $x$ to get $x^{2n+1}$. Then, with $x=\frac{1}{2}$ I should get the sum I was looking for. Right?

Well, I've went through all my steps, did the differentiation a couple of times and confirmed it was correct with Wolfram Alpha but my end result is

$-\ln\sqrt{5}$ while Wolfram says it should be $\ln{\frac{2}{\sqrt{5}}}$

What did I do wrong?

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The sum you used starts at $n=0$ whereas the one you want starts at $n=1$.

Also recall that $-\ln(x)=\ln\left(\frac{1}{x}\right)$

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$$\sum_{n=1}^\infty \frac{(-1)^nx^n}{n}=-\log(1+x)$$ so with $x=\frac{1}{4}$ we have $$\frac{1}{2}\sum_{n=1}^\infty \frac{(-1)^nx^n}{n}=\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n\times2^{2n+1}}=-\frac{1}{2}\log(\frac{5}{4})=\log(\frac{2}{\sqrt{5}})$$

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Looks good to me! –  amWhy Jun 24 at 15:21

The détours you make in your solution are a bit frightening, I must say... One might rather start from the series $S(t)=\sum\limits_{n\geqslant1}\frac{t^n}n$, whose sum is $S(t)=-\log(1-t)$ for every $|t|\lt1$. Plugging $t=-\frac14$ in this shows that the sum of the series in your post is $\frac12S\left(-\frac14\right)=-\frac12\log\frac54=\log\left(\frac2{\sqrt5}\right)$.

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