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$$\sum_{n=1}^{\infty} \frac{1}{n 2^{2n+1}} = \ln \left(\frac{2}{\sqrt{3}}\right)$$

I could show convergence. (I dont need to show that this converges). However I couldn't figure how to show the value.

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1  
Did you try $\sum_{n=1}^{\infty} \frac{y^n}{n} = -\ln \left(1-y\right)$ –  Kirthi Raman May 23 '12 at 12:53
    
"I dont need to show that this converges" implies it is a homework in my opinion. –  Wok May 23 '12 at 12:57
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@wok adding the homework tag to something that is not obviously homework is considered impolite. This could be a question from a textbook. –  Chris Taylor May 23 '12 at 12:59
    
I'm going to delete the homework tag. Ehrlick, if it is homework, please add the homework tag back in. –  Gerry Myerson May 23 '12 at 13:01
    
@The-Ever-Kid, we don't add the homework tag to other people's questions unless the person asking the question says that it's homework. We encourage the poster to add the tag, if it is homework. –  Gerry Myerson May 24 '12 at 13:49

3 Answers 3

$$ \begin{align*} \sum_{n=1}^{\infty} \frac{y^n}{n} &= -\ln \left(1-y\right) \hspace{15pt} {\textit{apply }} \hspace{5pt} y=\frac{1}{x^2}\\ \sum_{n=1}^{\infty} \frac{1}{n x^{2n}} &= -\ln \left(1-\frac{1}{x^2}\right) \\ \sum_{n=1}^{\infty} \frac{1}{n 2^{2n+1}} &= \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n 2^{2n}} = -\frac{1}{2}\ln \left(1-\frac{1}{4}\right)\\ \end{align*} $$

Do the rest to simplify and get what you want.

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$$ f(x) = \sum_{n=1}^{\infty} \frac{x^n}{n} \Rightarrow f'(x) = \sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x} $$ $$ f(x) = \int \frac{1}{1-x} \ dx = -\ln(1-x) + c $$ $$ but : f(0) = 0 \Rightarrow 0 + c = 0 \Rightarrow c = 0 \Rightarrow f(x) = \sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln(1-x) $$ –  what'sup Aug 13 '13 at 15:21
    
@what'sup An alternative derivation: Start with$$\lim_{\epsilon\to0}\dfrac{(1-x)^\epsilon-1}\epsilon$$which L'Hôpital shows is equal to $\ln(1-x)$. Then, use Newton's generalized binomial theorem to expand that into a series. –  columbus8myhw Oct 12 at 17:23

There are various options - here is one: I won't work it through to the end.

Write $$f(t)=\sum_{n=1}^{\infty} \frac{t^n}{n 2^{2n+1}}$$

Then $$f'(t)=\sum_{n=1}^{\infty} \frac{t^{n-1}}{2^{2n+1}} = \frac 1 8 \sum_{n=1}^{\infty} \left(\frac t 4\right)^{n-1}$$

Which is a geometric series ...

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Good answer! You have to be a little bit careful about the domain of definition of $f$ (because the radius of convergence of $f$ is 4, by the ratio test) but as we are only interested in $t=1$ this is of course no problem. –  Nils Matthes May 23 '12 at 13:13

Artin's comment pretty much answers the question.

Did you try $\sum_{n=1}^\infty\frac{y^n}{n}=-ln(1-y)$

Just observe that your series actually is $$\frac{1}{2}\sum_{n=1}^\infty\frac{\left(1/4\right)^n}{n}$$ As this is homework, you should also, I guess, be careful to check radius of convergence...

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$$ f(x) = \sum_{n=1}^{\infty} \frac{x^n}{n} \Rightarrow f'(x) = \sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x} $$ $$ f(x) = \int \frac{1}{1-x} \ dx = -\ln(1-x) + c $$ $$ but : f(0) = 0 \Rightarrow 0 + c = 0 \Rightarrow c = 0 \Rightarrow f(x) = \sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln(1-x) $$ –  what'sup Aug 13 '13 at 15:21

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