<s>[INST] Summing a given series 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: $$S = 1 + \Bigl(1 + \frac{1}{3}\Bigr) \cdot \frac{1}{5} \cdot \frac{1}{3} + \Bigl(1 + \frac{1}{3} + \frac{1}{5}\Bigr)\cdot \frac{1}{5^{2}} \cdot \frac{1}{5} + \Bigl(1 + \frac{1}{3}+ \frac{1}{5} + \frac{1}{7}\Bigr) \cdot \frac{1}{5^{3}} \cdot \frac{1}{7} + \cdots + \text{ad inf}$$
 [/INST] 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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