How to calculate this sum: $\sum_{k=1}^{n} \frac{k}{n^k} \binom nk$

Question

How do you calculate this sum

$$ \sum \limits_{k=1}^{n} \frac{k}{n^k}{n\choose k} \;?$$

Answer 1

Since $n-k = (n-1) - (k-1)$, $$ \sum_{k=1}^n \frac{k}{n^k}\binom{n}{k} = \sum_{k=1}^n \frac{1}{n^{k-1}}\frac{(n-1)!}{(k-1)!(n-k)!} = \sum_{k=1}^n \left(\frac{1}{n}\right)^{k-1} \binom{n-1}{k-1} $$ now do a change of variable to get, by the binomial expansion, $$ \sum_{k=1}^n \frac{k}{n^k}\binom{n}{k} = \sum_{k=0}^{n-1} \left(\frac{1}{n}\right)^k \binom{n-1}{k} = \left(1 + \frac{1}{n}\right)^{n-1} $$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{k = 1}^{n}{k \over n^{k}}{n\choose k}:\ {\large ?}}$

\begin{align} \color{#00f}{\large\sum_{k = 1}^{n}{k \over n^{k}}{n\choose k}}&= \left. x\,\partiald{}{x}\sum_{k = 1}^{n}{n\choose k}x^{k}\right\vert_{x\ =\ 1/n} =\left. x\,\partiald{\pars{1 + x}^{n}}{x}\right\vert_{x\ =\ 1/n} =\left. \vphantom{\huge A}x\,n\pars{1 + x}^{n - 1}\right\vert_{x\ =\ 1/n} \\[3mm]&=\color{#f00}{1 \over n}\,n\pars{1 + \color{#f00}{1 \over n}}^{n - 1} =\color{#00f}{\large\pars{1 + {1 \over n}}^{n - 1}} \end{align}