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I got this problem from one of my mates, and i rearranged them, and got it in a summable form. From here could anyone tell me as to how i can sum up this interesting series.

How does one sum the given series: $S = 1 + \Bigl(1 + \frac{1}{3}\Bigr) \cdot \frac{1}{5} \cdot \frac{1}{3} + \Bigl(1 + \frac{1}{3} + \frac{1}{5}\Bigr)\cdot \frac{1}{5^{2}} \cdot \frac{1}{5} + \Bigl(1 + \frac{1}{3}+ \frac{1}{5} + \frac{1}{7}\Bigr) \cdot \frac{1}{5^{3}} \cdot \frac{1}{7} + \cdots + \text{ad inf}$

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I presume this is $\sum_{n=0}^\infty a_n 5^{-n}$ where $(2n+1)a_n=\sum_{k=0}^n 1/(2k+1)$. Let $f(x)=\sum_{n=0}^\infty a_n x^{2n+1}$ so that $f(0)=0$ and $f'(x)=\sum_{n=0}^\infty x^{2n}\sum_{k=0}^n\frac{1}{2k+1}.$ Therefore $f'(x)=\frac{1}{1-x^2}\frac{\tanh^{-1}x}{x}.$ Your sum equals $\int_0^{1/5}\frac{\tanh^{-1}x}{x(1-x^2)}dx.$ If we let $x=\tanh y$ then $dx=(1-x^2)dy$ and we get $\int_0^t y\coth y\ dy$ where $t=\tanh^{-1}(1/5)$. I'm getting a nasty feeling that $\int y\coth y\ dy$ is one of those integrals that can't be done elementarily. :-(

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    Even thought the indefinite integral is hard, there might be ways to find the definite integral...2010-11-03
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$S(x) = 1 + \Bigl(1 + \frac{1}{3}\Bigr) \cdot \frac{x^2}{5} \cdot \frac{1}{3} + \Bigl(1 + \frac{1}{3} + \frac{1}{5}\Bigr)\cdot \frac{x^4}{5^{2}} \cdot \frac{1}{5} + \Bigl(1 + \frac{1}{3}+ \frac{1}{5} + \frac{1}{7}\Bigr) \cdot \frac{x^6}{5^{3}} \cdot \frac{1}{7} + \cdots$ $=\sum_{k=0}^{\infty} \Bigl(1 + \frac{1}{3}+\cdots+\frac{1}{2k+1}\Bigr) \cdot \left(\frac{x}{\sqrt 5}\right)^{2k} \cdot \frac{1}{2k+1}$ $=\frac{1}{x}\int_{0}^x\sum_{k=0}^{\infty} \Bigl(1 + \frac{1}{3}+\cdots+\frac{1}{2k+1}\Bigr) \cdot \left(\frac{y}{\sqrt 5}\right)^{2k}dy$ $=\frac{\sqrt 5}{x}\int_{0}^{x/\sqrt 5}\sum_{k=0}^{\infty} \Bigl(1 + \frac{1}{3}+\cdots+\frac{1}{2k+1}\Bigr) t^{2k+1}\frac{dt}{t}.$

Now let $F(t)=\sum_{k=0}^{\infty} \Bigl(1 + \frac{1}{3}+\cdots+\frac{1}{2k+1}\Bigr) t^{2k+1},$ then $F(t)+\frac{t}{2}F(t)=H(t):= \sum_{n=0}^{\infty} \Bigl(1 + \frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\Bigr) t^n.$ $H(t)$ is the generating function of the harmonic numbers and is well known: $H(t)=\frac{-\ln(1-t)}{1-t}.$

Combining it all together we get that $S(x)=\frac{\sqrt 5}{x}\int_{0}^{x/\sqrt 5} \frac{2\ln(1-t)}{t(t-1)(t+2)}dt$ which probably can be calculated in elementary functions.

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    C$a$n i get a reference $b$y which I can get used to these type of Summations. You might have seen them.2011-05-16
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HINT $\ $ If you massage the odd bisection of the generating function of the harmonic numbers then you should obtain a closed form in terms of (di)logs evaluated in $\:\mathbb Q(\sqrt{5})$

EDIT $\ $ See Andrey's later answer for some further details of the method I hinted above.

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    @Andrey: No problem. Surely it will help others to see the details fully worked out.2010-11-03