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: $$S = 1 + \Bigl(1 + \frac{1}{3}\Bigr) \cdot \frac{1}{5} \cdot \frac{1}{3} + \Bigl(1 + \frac{1}{3} + \frac{1}{5}\Bigr)\cdot \frac{1}{5^{2}} \cdot \frac{1}{5} + \Bigl(1 + \frac{1}{3}+ \frac{1}{5} + \frac{1}{7}\Bigr) \cdot \frac{1}{5^{3}} \cdot \frac{1}{7} + \cdots + \text{ad inf}$$
 A: 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. :-(
A: $$S(x) = 1 + \Bigl(1 + \frac{1}{3}\Bigr) \cdot \frac{x^2}{5} \cdot \frac{1}{3} + \Bigl(1 + \frac{1}{3} + \frac{1}{5}\Bigr)\cdot \frac{x^4}{5^{2}} \cdot \frac{1}{5} + \Bigl(1 + \frac{1}{3}+ \frac{1}{5} + \frac{1}{7}\Bigr) \cdot \frac{x^6}{5^{3}} \cdot \frac{1}{7} + \cdots$$
$$=\sum_{k=0}^{\infty} \Bigl(1 + \frac{1}{3}+\cdots+\frac{1}{2k+1}\Bigr) \cdot \left(\frac{x}{\sqrt 5}\right)^{2k} \cdot \frac{1}{2k+1}$$
$$=\frac{1}{x}\int_{0}^x\sum_{k=0}^{\infty} \Bigl(1 + \frac{1}{3}+\cdots+\frac{1}{2k+1}\Bigr) \cdot \left(\frac{y}{\sqrt 5}\right)^{2k}dy$$ 
$$=\frac{\sqrt 5}{x}\int_{0}^{x/\sqrt 5}\sum_{k=0}^{\infty} \Bigl(1 + \frac{1}{3}+\cdots+\frac{1}{2k+1}\Bigr) t^{2k+1}\frac{dt}{t}.$$
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.
A: 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.
