A double sum

[akotronis]

For \(x\in(-1,1]\), calculate \(\displaystyle\sum_{n\geq0}(-1)^n\left(\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}+\frac{x^{n+3}}{n+3}-\cdots\right)\).

Proposed by Ovidiu Furdui.

[Demetres]

For \(|x| < 1\) observe that \[ \int_0^x \frac{t^n}{1+t} \, dt = \int_0^x \sum_{k=0}^{\infty} (-1)^k t^{n+k} \, dt = \sum_{k=0}^{\infty} \int_0^x (-1)^k t^{n+k} \, dt = \sum_{k=0}^{\infty} \frac{(-1)^k x^{n+k+1}}{n+k+1}. \] Here, the interchange of the limiting operations is allowed because the convergence is absolute. For \(x=1\) the convergence is not absolute but taking the limit as \(x\) tends to \(1\) by the monotone convergence theorem we also have that \[ \int_0^1 \frac{t^n}{1+t} \, dt = \sum_{k=0}^{\infty} \frac{(-1)^k}{n+k+1}.\] So the original sum is equal to \[ \sum_{n=0}^{\infty}(-1)^n\int_0^x \frac{t^n}{1+t} \, dt = \int_0^x \sum_{n=0}^{\infty} \frac{(-t)^n}{1+t} \, dt = \int_0^x \frac{1}{(1+t)^2} \, dt = 1 - \frac{1}{x+1} = \frac{x}{x+1}. \] Here the interchange of the limiting operations is allowed by the dominated convergence theorem as \[ f_n(t) = \sum_{k=0}^n \frac{(-t)^k}{1+t}\] is dominated by \(\frac{2}{(1+t)^2}.\)

[akotronis]

Good morning. The problem is from here

The solution I had sent: (It uses DCT too)

\begin{align}\sum_{n\geq0}(-1)^n\sum_{k\geq1}(-1)^{k-1}\frac{x^{n+k}}{n+k}&=\sum_{n\geq0}\sum_{k\geq1}(-1)^{n+k-1}\frac{x^{n+k}}{n+k}\notag \\ &\stackrel{m:=n+k}{=}\sum_{n\geq0}\sum_{m\geq n+1}(-1)^{m-1}\frac{x^{m}}{m}\notag \\ &\stackrel{x\in(-1,1]}{=}\sum_{n\geq0}\left(\ln(1+x)-\sum_{m=1}^{n}(-1)^{m-1}\frac{x^m}{m}\right)\notag \\ &=\lim_{N\to+\infty}\sum_{n=0}^{N}\left(\ln(1+x)-\sum_{m=1}^{n}(-1)^{m-1}\frac{x^m}{m}\right)\notag \\ :&=\lim_{N\to+\infty}A_N(x).\notag\end{align}

Now for \(x\in(-1,1]\) it is

\[A^{'}_N(x)=\sum_{n=0}^{N}\left(\frac{1}{1+x}-\sum_{m=1}^{n}(-x)^{m-1}\right)=\sum_{n=0}^{N}\frac{(-x)^n}{1+x}=\frac{1-(-x)^{N+1}}{(1+x)^2}\]

so integrating we get

\[A_N(x)=\int_{0}^{x}\frac{1-(-y)^{N+1}}{(1+y)^2}\,dy=\frac{x}{1+x}+(-1)^N\int_{0}^{x}\frac{y^{N+1}}{(1+y)^2}\,dy.\]

But for \(x\in(-1,1)\) it is

\[\Big|\int_{0}^{x}\frac{y^{N+1}}{(1+y)^2}\,dy\Big|\leq\int_{0}^{|x|}\frac{y^{N+1}}{(1-y)^2}\,dy\leq|x|^{N+1}\int_{0}^{|x|}\frac{1}{(1-y)^2}\,dy=\frac{|x|^{N+2}}{1-|x|}\to0\]

and for \(x=1\), \(\displaystyle\int_{0}^{1}\frac{y^{N+1}}{(1+y)^2}\,dy\to0\) by the Dominated Convergence theorem, so

\[\sum_{n\geq0}(-1)^n\sum_{k\geq1}(-1)^{k-1}\frac{x^{n+k}}{n+k}=\frac{x}{x+1},\qquad x\in(-1,1].\]