How to find the sum of the following series:
$$1+\frac13\frac14+\frac15\frac1{4^2}+\dots?$$ The general term is $u_n=\frac1{(2n+1)4^n};n\geq 1$.
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1If $u_n=\frac{1}{2n+1}\frac{1}{2^n}$ then the sum is $$1+\frac{1}{3}\frac{1}{2}+\frac{1}{5}\frac{1}{2^2}+\ldots$$ – Ángel Mario Gallegos Jan 21 '15 at 16:18
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Do you know how to find closed forms of series like $\displaystyle \sum\limits_{n=1}^{\infty} \frac{x^{2n+1}}{2n+1}$ for $|x| < 1$? – sciona Jan 21 '15 at 16:18
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It appears the general term is $u_n=\frac 1{2n+1}\frac 1{4^n}$ – Ross Millikan Jan 21 '15 at 16:21
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Wolfram gives the result as this – Varun Iyer Jan 21 '15 at 16:25
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HINT:
$$\ln(1+x)=x-\dfrac{x^2}2+\dfrac{x^3}3-\dfrac{x^4}4+\cdots$$
$$\ln(1-x)=-x-\dfrac{x^2}2-\dfrac{x^3}3-\dfrac{x^4}4-\cdots$$
$$\ln(1+x)-\ln(1-x)=?$$
Keep in mind : Taylor series for $\log(1+x)$ and its convergence
lab bhattacharjee
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How did you get the idea that the series would be this one as there is no general technique to view that – Learnmore Jan 22 '15 at 03:29
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@learnmore, Whenever you have something like $$\frac{x^n}n$$ think of log series for $$\frac{x^n}{n!}$$ think of exponential series. For $$\frac{x^n}{r(r+1)\cdots(r+n-1)}$$ think of http://en.wikipedia.org/wiki/Binomial_series – lab bhattacharjee Jan 22 '15 at 04:47
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By using $\frac{4}{4-x^2}=\frac{1}{2-x}+\frac{1}{2+x}$ we have: $$\sum_{n\geq 0}\frac{1}{(2n+1)4^n}=\sum_{n\geq 0}\int_{0}^{1}\frac{x^{2n}}{4^n}\,dx = \int_{0}^{1}\frac{4\,dx}{4-x^2}=\color{red}{\log 3}.$$
Jack D'Aurizio
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not so sure how you are removing the summation .Would have been nice if you could explain a bit – Learnmore Jan 22 '15 at 02:55
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\begin{align*} \sum_{n=0}^{\infty}\frac{1}{2n+1}\frac{1}{4^n}&=2\sum_{n=0}^{\infty}\frac{1}{2n+1}\left(\frac{1}{2}\right)^{2n+1}\\ &=2\,\text{artanh}\left(\frac{1}{2}\right) \end{align*} Since arctanh ($x$) can be expanded as $\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}$ for $|x|<1$.
Ángel Mario Gallegos
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