How to Find closed form $\sum_{n=1}^{\infty}\frac{x^{2n}}{n}a_n$

Question

How to Find closed form :$$\sum_{n=1}^{\infty}\frac{x^{2n}}{n}a_n$$ where $a_n=\sum_{k=1}^{2n-1}\frac{(-1)^{k-1}}{k}$

$$S(x)=\sum_{n=1}^{\infty}\frac{x^{2n}}{n}a_n=\sum_{n=1}^{\infty}\int^1_0y^{n-1}x^{2n}a_n dy =\int^1_0 \frac{1}{y}\sum_{n=1}^{\infty}a_n(yx^2)^n dy =x^2\int^1_0 \sum_{n=1}^{\infty}a_{n+1}(yx^2)^n dy$$

we have : $$\boxed{\sum_{n=1}^{\infty}a_nx^n=\dfrac{1}{1-x}\sum_{n=1}^{\infty}(a_n-a_{n-1})x^n}$$

therfore: $$S(x)=x^2\int^1_0 \frac{1}{1-yx^2}\sum_{n=1}^{\infty}(a_{n+1}-a_n)(yx^2)^n dy=\int^1_0\frac{-x^2}{1-yx^2}\sum_{n=1}^{\infty}\frac{(yx^2)^n}{2n(2n+1)}dy$$ Let :$$S_1=\sum_{n=1}^{\infty}\frac{(yx^2)^n}{2n(2n+1)}=\frac{1}{x\sqrt y}\sum_{n=1}^{\infty}\frac{(x\sqrt y)^{2n+1}}{2n(2n+1)}$$ let :$t:=x\sqrt y$ therfore: $$S_1=\frac{1}{2t}\int^t_0\sum_{n=1}^{\infty}\frac{(z^2)^n}{n} dz =-\frac{1}{2t}\int^t_0\ln(1-z^2)dz=-\frac{1}{2t}\left({t\ln(1-t^2)-2t+\ln(1+t)-\ln(1-t)}\right)$$ Then: $$S_1=1-\frac{1}{2}\ln(1-yx^2)-\frac{1}{2x \sqrt y}\ln(1+x \sqrt y)+\frac{1}{2x \sqrt y}\ln(1-x \sqrt y)$$

we have : $$S(x)=\int^1_0\frac{-x^2}{1-yx^2}S_1dy=I_1+I_2+I_3+I_4$$

where : $$I_1=\int^1_0\frac{-x^2}{1-yx^2}dy=\ln(1-x^2)$$ $$I_2=\frac{1}{2}\int^1_0\frac{x^2\ln(1-yx^2)}{1-yx^2}dy=-\frac{1}{4}\ln^2(1-x^2)$$ $$I_3=\frac{1}{2}\int^1_0\frac{x\ln(1+x \sqrt y)}{ \sqrt y(1-yx^2)}dy$$ $$I_4=-\frac{1}{2}\int^1_0\frac{x\ln(1-x \sqrt y)}{ \sqrt y(1-yx^2)}dy$$

How To Evaluate $I_3$ and $I_4$ ??

Is there a better way than this??

Thank you very much for your interest

Answer 1

\begin{align*} \sum_{n=1}^\infty\frac{x^{2n}}n\sum_{k=1}^{2n-1}\frac{(-1)^{k-1}}k &=\sum_{n=1}^\infty\frac{x^{2n}}{2n}\sum_{k=1}^{2n-1}\left(\frac{(-1)^{k-1}}{k}+\frac{(-1)^{2n-k-1}}{2n-k}\right) \\&=\sum_{n=1}^\infty x^{2n}\sum_{k=1}^{2n-1}\frac{(-1)^{k-1}}{k(2n-k)}=\frac12\big(f(x)+f(-x)\big), \end{align*} where $$ f(x)=\sum_{n=2}^\infty x^n\sum_{k=1}^{n-1}\frac{(-1)^{k-1}}{k(n-k)}=\color{blue}{-\log(1+x)\log(1-x)}=f(-x) $$ (as a Cauchy product). Hence the result is $f(x)$.

Answer 2

Here's how I'd get related integrals without the $y$ and $z$:

Note $a_n$ converges to $-\ln 2$, so by the ratio test $S(x)$ converges if $|x|<1$.

$$ S(x) = \sum_{n=1}^\infty \frac{x^{2n}}{n} \left[\sum_{k=1}^{2n-1} \frac{(-1)^{k-1}}{k}\right] $$

$$ S(x) = \sum_{n=1}^\infty \sum_{k=1}^{2n-1} \frac{(-1)^{k-1}}{nk} x^{2n} $$

$$ S'(x) = 2 \sum_{n=1}^\infty \sum_{k=1}^{2n-1} \frac{(-1)^{k-1}}{k} x^{2n-1} $$

The set of pairs $(n,k)$ in the sum are described by $n \geq 1$ and $1 \leq k \leq 2n-1$ or equivalently, $k \geq 1$ and $n \geq \frac{k+1}{2}$. Reordering the sums,

$$ S'(x) = 2 \sum_{k=1}^\infty \sum_{n= \left\lceil \frac{k+1}{2} \right\rceil}^\infty \frac{(-1)^{k-1}}{k} x^{2n-1} $$

$$ S'(x) = 2 \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} \cdot \frac{x^{2\lceil (k+1)/2 \rceil-1}}{1-x^2} $$

Rewrite this in terms of its odd terms with $k=2\ell-1$ and even terms with $k=2\ell$:

$$ S'(x) = \frac{2}{1-x^2} \sum_{\ell=1}^\infty \left[\frac{x^{2\ell-1}}{2\ell-1} - \frac{x^{2\ell+1}}{2\ell} \right] $$

If $f(t) = \sum_{k=1}^\infty \frac{t^k}{k} = -\ln(1-t)$ for $|t|<1$, then

$$ \sum_{\ell=1}^\infty \frac{x^{2\ell-1}}{2\ell-1} = \frac{f(x)-f(-x)}{2} = -\frac 12 \ln(1-x) + \frac 12 \ln(1+x) $$

$$ \sum_{\ell=1}^\infty \frac{x^{2\ell}}{2\ell} = \frac{f(x)+f(-x)}{2} = - \frac 12 \ln(1-x) - \frac 12 \ln(1+x)$$

$$ S'(x) = \frac{(-1+x)\ln(1-x)+(1+x)\ln(1+x)}{1-x^2} $$

$$ S'(x) = -\frac{1}{1+x}\ln(1-x) + \frac{1}{1-x}\ln(1+x) $$

This has a derivative product rule form:

$$ S(x) = -\ln(1-x) \ln(1+x) + C$$

Since the definition gives us $S(0)=0$, $C=0$.

Answer 3

Probably too advanced but it is too long for a comment.

The problem is interesting since, for once, integration is not required $$a_n=\sum_{k=1}^{2n-1}\frac{(-1)^{k-1}}{k}=\log (2)+\Phi (-1,1,2 n)$$ where appears the Lerch transcendent function. $$\sum_{n=1}^{\infty}\frac{x^{2n}}{n}a_n=\log(2)\sum_{n=1}^{\infty}\frac{x^{2n}}{n}+\sum_{n=1}^{\infty}\frac{x^{2n}}{n}\Phi (-1,1,2 n)$$ $$\sum_{n=1}^{\infty}\frac{x^{2n}}{n}a_n=-\log (2) \log \left(1-x^2\right)+\Big(\log (2) \log \left(1-x^2\right)-\log (1-x) \log (1+x) \Big)$$ $$\sum_{n=1}^{\infty}\frac{x^{2n}}{n}a_n=-\log (1-x) \log (1+x)$$

You could even play with the partial sums since, as an approximation, $$\Phi (-1,1,2 n)\sim \frac 1 {4n}$$