MHB What is the Interesting Euler Sum Proven by this Equation?

[alyafey22]

Prove the following Euler sum

$$\sum_{k\geq 1}\left(1+\frac{1}{3}+\cdots +\frac{1}{2k-1} \right) \frac{x^{2k}}{k}=\frac{1}{4}\ln^2\left( \frac{1+x}{1-x}\right)$$​

 

[polygamma]

I get that

$$\sum_{k=1}^{\infty} H_{2k-1} \frac{x^{2k}}{k} = \frac{1}{2} \Big( \ln^{2}(1-x) + \ln^{2}(1+x) \Big)$$

Spoiler

The generating function for the harmonic numbers is $ \displaystyle -\frac{\ln(1-x)}{1-x} $.

So

$$ \sum_{k=1}^{\infty} (-1)^{k} H_{k} x^{k} = - \frac{\ln(1+x)}{1+x}$$

And

$$ \sum_{k=1}^{\infty} H_{k} x^{k} - \sum_{k=1}^{\infty} (-1)^{k} H_{k} x^{k} = 2 \sum_{k=1}^{\infty} H_{2k-1} x^{2k-1} = -\frac{\ln(1-x)}{1-x} + \frac{\ln(1+x)}{1+x}$$

Now integrate.

$$

\sum_{k=1}^{\infty} H_{2k-1} \frac{x^{2k}}{k} = - \int \frac{\ln(1-x)}{1-x} \ dx + \int \frac{\ln(1+x)}{1+x} \ dx $$

$$ = \frac{1}{2} \Big( \ln^{2}(1-x) + \ln^{2}(1+x) \Big) + C$$

And when $x$ is zero, $ \displaystyle \frac{1}{2} \Big( \ln^{2}(1-x) + \ln^{2}(1+x) \Big) =0$. So $C = 0 $.

 

[alyafey22]

I think you should consider the harmonic numbers

$$H_{2k}+ H'_{2k}$$ , that's way you got a different answer .

 

[polygamma]

But our answers aren't equivalent.

 

[alyafey22]

$$1+\frac{1}{3}+\frac{1}{5}+\cdots +\frac{1}{2k-1}= \left(1+\frac{1}{2}+\frac{1}{3} +\cdots +\frac{1}{2k-1}+\frac{1}{2k}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{k} \right)=H_{2k} -\frac{1}{2}H_k$$

- - - Updated - - -

I think you are evaluating $$H_{2k-1}\neq 1+\frac{1}{3}+\frac{1}{5}+\cdots +\frac{1}{2k-1}$$

 

[polygamma]

Yeah. My mistake. Sorry.

 

[polygamma]

Redo (Smile)

Spoiler

$ \displaystyle \sum_{k =1}^{\infty} \left(1+\frac{1}{3}+\cdots +\frac{1}{2k-1} \right) \frac{x^{2k}}{k} = \sum_{k=1}^{\infty} \left( H_{2k} - \frac{1}{2} H_{k} \right) \frac{x^{2k}}{k} $$\displaystyle \sum_{k=1}^{\infty} H_{k} x^{k} + \sum_{k=1}^{\infty} (-1)^{k} H_{k} x^{k} = 2 \sum_{k=1}^{\infty} H_{2k} x^{2k} = - \frac{\ln(1-x)}{1-x} - \frac{\ln(1+x)}{1+x}$$\displaystyle \sum_{k=1}^{\infty} H_{2k} \frac{x^{2k}}{k} = - \frac{1}{2} \int \frac{\ln(1-x)}{x(1-x)} \ dx - \frac{1}{2} \int \frac{\ln(1+x)}{x(1+x)} \ dx $

$ \displaystyle = \text{Li}_{2} (x) + \frac{1}{2} \ln^{2}(1-x) + \text{Li}_{2}(-x) + \frac{1}{2} \ln^{2}(1+x) + C $ (partial fractions)

$ \displaystyle = \frac{1}{2} \text{Li}_{2}(x^{2}) + \frac{1}{2} \ln^{2}(1-x) + \frac{1}{2} \ln^{2}(1+x) + C$The constant of integration must be zero.$ \displaystyle \sum_{k=1} H_{k} \frac{x^{2k}}{k} = \text{Li}_{2}(x^{2})+ \frac{1}{2} \ln^{2}(1-x^{2})$So

$ \displaystyle \sum_{k=1}^{\infty} \left( H_{2k} - \frac{1}{2} H_{k} \right) \frac{x^{2k}}{k}= \frac{1}{2} \text{Li}_{2}(x^{2}) + \frac{1}{2} \ln^{2}(1-x) + \frac{1}{2} \ln^{2}(1+x) - \frac{1}{2} \text{Li}_{2}(x^{2}) - \frac{1}{4} \ln^{2}(1-x^{2}) $

$ \displaystyle = \frac{1}{2} \ln^{2}(1-x) + \frac{1}{2} \ln^{2}(1+x) - \frac{1}{4} \Big( \ln(1-x) + \ln(1+x) \Big)^{2} $

$ \displaystyle = \frac{1}{4} \ln^{2}(1-x) + \frac{1}{4} \ln^{2}(1+x) - \frac{1}{2} \ln(1-x) \ln(1+x)$

$ \displaystyle = \Big( \frac{1}{2} \ln(1-x) - \frac{1}{2} \ln(1+x) \Big)^{2} $

$ \displaystyle = \frac{1}{4} \ln^{2} \Big( \frac{1+x}{1-x} \Big) $

 

[alyafey22]

$\displaystyle \sum_{k=1}^{\infty} H_{k} x^{k} + \sum_{k=1}^{\infty} (-1)^{k} H_{k} x^{k} = 2 \sum_{k=1}^{\infty} H_{2k} x^{2k} = \cdots $

I am not convinced of this step , how did you get $$H_{2k}$$ ?

 

[polygamma]

$ \displaystyle \sum_{k=1}^{\infty} H_{k} x^{k} + \sum_{k=1}^{\infty} (-1)^{k} H_{k} x^{k} = (H_{1} x + H_{2} x^{2} + H_{3}x^{3}+ H_{4}x^{4} \ldots) + (-H_{1} x^{1} + H_{2} x^{2} - H_{3}x^{3} + H_{4}x^{4} \ldots ) $

$ \displaystyle = 2 H_{2}x^{2} + 2 H_{4}x^{4} + \ldots $

$ \displaystyle = 2 \sum_{k=1}^{\infty} H_{2k} x^{2k} $