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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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3 Answers 3

up vote 6 down vote accepted

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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2  
The last integral has a closed form in terms of (di)logs -which jibes with my intuition as to what would result from the generating-function approach I hinted at. –  Bill Dubuque Nov 3 '10 at 19:40
    
Even thought the indefinite integral is hard, there might be ways to find the definite integral... –  Aryabhata Nov 3 '10 at 22:03

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

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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No, I think as Bill Dubuque says you should get dilogarithms. (Take partial fractions to see it.) –  Qiaochu Yuan Nov 3 '10 at 19:54
    
Yes, that's precisely as I hinted. The integral has a closed form in terms of (di)logs - presuming your calculations are correct. –  Bill Dubuque Nov 3 '10 at 19:57
    
@Bill Dubuque and Qiaochu Yuan: Thanks for the comments! Now I see that my conclusion was too hasty. –  Andrey Rekalo Nov 3 '10 at 20:14
    
Can i get a reference by which I can get used to these type of Summations. You might have seen them. –  user9413 May 16 '11 at 15:50

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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I posted my answer before I noticed yours, sorry. –  Andrey Rekalo Nov 3 '10 at 20:20
    
@Andrey: No problem. Surely it will help others to see the details fully worked out. –  Bill Dubuque Nov 3 '10 at 20:25

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