Find limit of sum

Question

I suspect that $\lim_{n \to \infty} \sum_{k = 0}^{n - 1}\frac{k}{a^k(n - k)} = 0$ for $a > 1$. I know that this product represents the taylor coefficients of $\frac{-ax\ln(1 - x)}{(a - x)^2}$ by the Cauchy product. Unfortunately the limit as $x \to 1^{-1}$ is not defined so I can't use Abel's theorem. How can I prove this?

Answer 1

The limit is zero as expected. First, write the sum as

$$ \sum_{k=0}^{n-1} \frac{k}{a^k(n-k)} = \sum_{k=0}^{\infty} \frac{k}{a^k(n-k)} \mathbf{1}_{\{k < n\}}$$

Since each term is uniformly bounded by

$$ \left| \frac{k}{a^k(n-k)}\mathbf{1}_{\{k < n\}} \right| \leq \frac{k}{a^k} $$

and $\sum_{k=0}^{\infty} \frac{k}{a^k} < \infty$, the dominated convergence theorem tells that

$$ \lim_{n\to\infty} \sum_{k=0}^{\infty} \frac{k}{a^k(n-k)} \mathbf{1}_{\{k < n\}} = \sum_{k=0}^{\infty} \lim_{n\to\infty} \frac{k}{a^k(n-k)} \mathbf{1}_{\{k < n\}} = \sum_{k=0}^{\infty} 0 = 0. $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\left.\lim_{n \to \infty}\sum_{k = 0}^{n - 1} {k \over a^{k}\pars{n - k}}\,\right\vert_{\ a\ >\ 1} = 0:\ {\large ?}}$.

\begin{align} &\left.\lim_{n \to \infty}\sum_{k = 0}^{n - 1} {k \over a^{k}\pars{n - k}}\,\right\vert_{\ a\ >\ 1} = \lim_{n \to \infty}\sum_{k = 1}^{n - 1}{k \over a^{k}\pars{n - k}} = \lim_{n \to \infty}\sum_{k = 1}^{n - 1}{n - k \over a^{n - k}\,k} \\[5mm] = &\ \lim_{n \to \infty}\pars{% {n \over a^{n}}\sum_{k = 1}^{n - 1}{a^{k} \over k} - {1 \over a^{n}}\sum_{k = 1}^{n - 1}a^{k}} = \lim_{n \to \infty}\pars{% {n \over a^{n}}\sum_{k = 1}^{n - 1}{a^{k} \over k} - {1 \over a^{n}}\,a\,{a^{n - 1} - 1 \over a - 1}} \\[5mm] = &\ \lim_{n \to \infty}\pars{% \bbox[#ffd,10px]{\ds{{n \over a^{n}}\sum_{k = 1}^{n - 1}{a^{k} \over k}}} - {1 - a^{1 - n} \over a - 1}}\label{1}\tag{1} \\[5mm] & \bbx{\mbox{Note that}\quad \left.\lim_{n \to \infty}{1 - a^{1 - n} \over a - 1} \,\right\vert_{\ a\ >\ 1} = {1 \over a - 1}}\label{2}\tag{2} \end{align}

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\begin{align} \bbox[#ffd,10px]{\ds{{n \over a^{n}}\sum_{k = 1}^{n - 1}{a^{k} \over k}}} & = {n \over a^{n}}\sum_{k = 1}^{n - 1}a^{k}\int_{0}^{1}t^{k - 1}\,\dd t = {n \over a^{n - 1}}\int_{0}^{1}\sum_{k = 1}^{n - 1}\pars{at}^{k - 1}\,\dd t \\[5mm] & = {n \over a^{n - 1}}\int_{0}^{1}{\pars{at}^{n - 1} - 1 \over at - 1}\,\dd t = {n \over a^{n}}\int_{0}^{a}{t^{n - 1} - 1 \over t - 1}\,\dd t \\[5mm] & = {n \over a^{n}}\int_{0}^{a}{\pars{a - t}^{n - 1} - 1 \over \pars{a - t} - 1}\,\dd t = {n \over a^{n}}\int_{0}^{a}{\exp\pars{\ln\pars{\pars{a - t}^{n - 1} - 1}} \over \pars{a - t} - 1}\,\dd t\label{3}\tag{3} \\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,&\ {n \over a^{n}}\pars{a^{n - 1} - 1}\, {a^{n - 1} - 1 \over a^{n - 2}\pars{n - 1}}\,{1 \over a - 1} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\large\to}\,\,\,\bbx{1 \over a - 1} \label{4}\tag{4} \end{align}

\eqref{3} evaluation involves the application of the Laplace Method.

Finally, with \eqref{1}, \eqref{2} and \eqref{4}: $$ \bbx{\left.\lim_{n \to \infty}\sum_{k = 0}^{n - 1} {k \over a^{k}\pars{n - k}}\,\right\vert_{\ a\ >\ 1} = 0} $$

Answer 3

Abel's theorem remarks says that as $$\lim_{x\to 1^-} \, \frac{-a x \log (1-x)}{(a-x)^2}=+\infty;\;a>1$$ then the series of coefficients diverges to $\infty$ therefore, according to some experiment I made with Mathematica, the series diverges

Answer 4

Yes. The limit is zero. The following is an elementary proof.

Write $\beta=a-1>0$. Then $a^{k}=(1+\beta)^{k}\geq\frac{k(k-1)}{2}\beta^{2}\geq\frac{k^{2}}{4}\beta^{2}$ for $k\geq2$. Now $$ \sum_{k=2}^{n-1}\frac{k}{a^{k}(n-k)}\leq\sum_{k=2}^{n-1}\frac{4}{k^{2}\beta^{2}}\cdot\frac{k}{(n-k)}=\frac{4}{\beta^{2}}\sum_{k=2}^{n-1}\frac{1}{k(n-k)}. $$ Observe that $$ \sum_{k=1}^{n-1}\frac{1}{k(n-k)}=\frac{1}{n}\sum_{k=1}^{n-1}\left(\frac{1}{k}+\frac{1}{n-k}\right)=\frac{2S_{n}}{n}, $$ where $S_{n}=\sum_{k=1}^{n-1}\frac{1}{k}$. Note that $$ S_{n}=1+\sum_{k=2}^{n-1}\frac{1}{k}\leq1+\sum_{k=2}^{n-1}\int_{k-1}^{k}\frac{1}{x}dx\leq1+\int_{1}^{n}\frac{dx}{x}=1+\ln n. $$ By squeeze theorem, $\lim_{n\rightarrow\infty}\frac{2S_{n}}{n}=0$. Hence, by squeeze theorem again, $\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}\frac{k}{a^{k}(n-k)}=0$ .