Does $\sum_{k=0}^{\infty} \sum_{n=1}^{\infty} x^nn^k =cx-x\ln(x)+x\ln(1-x) $ for $0 < x < 1$ for some real $c$? If so, what is $c$?

Question

Let $g_k(x) =\sum_{n=1}^{\infty} x^nn^k $ and $G(x) =\sum_{k=0}^{\infty} g_k(x) =\sum_{k=0}^{\infty} \sum_{n=1}^{\infty} x^nn^k $.

For $0 < x < 1$ is $G(x) =cx-x\ln(x)+x\ln(1-x) $ for some real $c$ and, if so, what is the value of $c$?

I am not sure that $G(x)$ converges for any $x$. The order of summation can certainly not be reversed.

I am confident of this analysis of the $g_k(x)$.

$\begin{array}\\ g_k'(x) &=\sum_{n=1}^{\infty} nx^{n-1}n^k\\ &=\dfrac1{x}\sum_{n=1}^{\infty} x^{n}n^{k+1}\\ &=\dfrac1{x}g_{k+1}(x)\\ \text{so}\\ g_{k+1}(x) &=xg_k'(x)\\ \end{array} $

From this I can show that $g_k(x) =\dfrac{xh_k(x)}{(1-x)^{k+1}} $ where $h_k(x)$ is a polynomial of degree $k-1$ with coefficients that appear to be the Eulerian numbers.

The following analysis assumes that $G(x)$ does converge and is term-by-term differentiable.

Therefore (and this may not be true because of convergence problems)

$\begin{array}\\ G'(x) &=\sum_{k=0}^{\infty} g_k'(x)\\ &=\sum_{k=0}^{\infty} \dfrac1{x}g_{k+1}(x)\\ &=\dfrac1{x}\sum_{k=0}^{\infty} g_{k+1}(x)\\ &=\dfrac1{x}\sum_{k=1}^{\infty} g_{k}(x)\\ &=\dfrac1{x}(-g_0(x)+\sum_{k=0}^{\infty} g_{k}(x))\\ &=\dfrac1{x}(-\dfrac{x}{1-x}+G(x))\\ &=-\dfrac1{1-x}+\dfrac1{x}G(x)\\ \text{so}\\ \dfrac{x}{1-x} &=G(x)-xG'(x)\\ \text{or}\\ \dfrac{1}{x(1-x)} &=\dfrac{G(x)-xG'(x)}{x^2}\\ &=-(G(x)x^{-1})'\\ \\ (G(x)x^{-1})' &=\dfrac{-1}{x(1-x)}\\ &=-\dfrac1{x}-\dfrac1{1-x}\\ G(x)x^{-1} &=c-\ln(x)+\ln(1-x) \qquad\text{for some } c\\ G(x) &=cx-x\ln(x)+x\ln(1-x)\\ \end{array} $

Answer 1

A formal consideration. In terms of the Eulerian polynomials $$ g_k (x) = \frac{{xA_k (x)}}{{(1 - x)^{k + 1} }}, $$ whence $$ G(x) = \sum\limits_{k = 0}^\infty {\frac{{xA_k (x)}}{{(1 - x)^{k + 1} }}} . $$ By the exponential generating function of the Eulerian polynomials $$ \frac{x}{{{\rm e}^{ - z} - x}} = \sum\limits_{k = 0}^\infty {\frac{{xA_k (x)}}{{(1 - x)^{k + 1} }}\frac{{z^k }}{{k!}}} . $$ Hence, formally, $$ \sum\limits_{k = 0}^\infty {\frac{{xA_k (x)}}{{(1 - x)^{k + 1} }}} = \int_0^{ + \infty } {\frac{{{\rm e}^{ - z}x }}{{{\rm e}^{ - z} - x}}{\rm d}z} = x\int_0^{ + \infty } {\frac{{{\rm d}z}}{{1 - x{\rm e}^z }}} = - x\log x + x\log (x - 1). $$

Answer 2

For $0 < x < 1$ and all $k \ge 0$ is $$ g_k(x) =\sum_{n=1}^{\infty} x^nn^k \ge \sum_{n=1}^{\infty} x^n = \frac{x}{1-x} $$ so that $$ G(x)=\sum_{k=0}^{\infty} g_k(x) $$ is not convergent (the terms $g_k(x)$ do not converge to zero). More precisely, $\sum_{k=0}^{\infty} g_k(x)$ diverges to $+\infty$.

Also $\sum_{k=0}^{\infty} \sum_{n=1}^{\infty} x^nn^k$ is a double series of nonnegative terms. If it were convergent (for some $x$) then it would be absolutely convergent, and the order of summation could be interchanged: $$ \sum_{k=0}^{\infty} \sum_{n=1}^{\infty} x^nn^k = \sum_{n=1}^{\infty}\sum_{k=0}^{\infty} x^nn^k = \sum_{n=1}^{\infty}x^n\sum_{k=0}^{\infty} n^k $$ But $\sum_{k=0}^{\infty} n^k$ diverges for all $n \ge 1$.