How to calculate $\lim_{n \to \infty} \sum_{k=1}^n\binom nk k!k\frac{1}{n^k}$?

Question

How to calculate this limit : $$\lim_{n\rightarrow +\infty}u_n$$ with : $$u_n=\sum_{k=1}^n\binom nk k!k\frac{1}{n^k}$$ We can write this : $$u_n=\sum_{k=1}^n\frac{n}{n}\times\frac{n-1}{n}\times\cdots\times\frac{n-k+1}{n}\times k$$ But I can't find the solution.

Answer 1

Here is a simplification of @Markus Scheuer's answer. Let

$$ a_k = \prod_{j=1}^{k} \frac{n-j+1}{n}. $$

Then by OP's observation,

$$ u_n = \sum_{k=1}^{n} k a_k = n \sum_{k=1}^{n} \left( 1 - \frac{n-k}{n} \right) a_k = n \sum_{k=1}^{n}(a_k - a_{k+1}) = n(a_1 - a_{n+1}) = n. $$

Answer 2

Hint: The number $u_n$ has a nice telescoping property. Let's for example consider the case $n=5$.

\begin{align*} \color{blue}{u_5}&=1+2\cdot\frac{4}{5}+3\cdot\frac{4}{5}\cdot\frac{3}{5}+4\cdot\frac{4}{5}\cdot\frac{3}{5}\cdot\frac{2}{5} +\color{blue}{5}\cdot\frac{4}{5}\cdot\frac{3}{5}\cdot\frac{2}{5}\cdot\color{blue}{\frac{1}{5}}\\ &=1+2\cdot\frac{4}{5}+3\cdot\frac{4}{5}\cdot\frac{3}{5} +(4+1)\cdot\frac{4}{5}\cdot\frac{3}{5}\cdot\frac{2}{5}\\ &=1+2\cdot\frac{4}{5}+3\cdot\frac{4}{5}\cdot\frac{3}{5} +\color{blue}{5}\cdot\frac{4}{5}\cdot\frac{3}{5}\cdot\color{blue}{\frac{2}{5}}\\ &=1+2\cdot\frac{4}{5}+(3+2)\cdot\frac{4}{5}\cdot\frac{3}{5}\\ &=1+2\cdot\frac{4}{5}+\color{blue}{5}\cdot\frac{4}{5}\cdot\color{blue}{\frac{3}{5}}\\ &=1+(2+3)\cdot\frac{4}{5}\\ &=1+\color{blue}{5}\cdot\color{blue}{\frac{4}{5}}\\ &=(1+4)\\ &\color{blue}{\,=5} \end{align*}

indicating $\color{blue}{u_n=n}$ for $n\geq 1$.

We observe we can iteratively collect the two right-most summands whereby the factor $5$ and $\frac{1}{5}$ cancel in the right-most summand.

Answer 3

Note that $$\begin{aligned} \sum_{k=1}^n\binom nk k!k\frac{1}{n^k} &= \sum_{k=1}^n\binom nk ((k+1)!-k!)\frac{1}{n^k}\\ &= \sum_{k=1}^n\binom nk (k+1)!\frac{1}{n^k}-\sum_{k=1}^n\binom nk k!\frac{1}{n^k}\\ &= \frac{n^2-n-1}{n+1}+\frac{e^n n^{-n} }{n+1}\Gamma (n+2,n) - \left(e^n n^{-n} \Gamma (n+1,n)-1 \right)\\ &= \frac{n^2-n-1}{n+1}+\frac{e^n n^{-n} }{n+1}\left((n+1)\Gamma (n+1,n) + n^{n+1}e^{-n}\right) - \left(e^n n^{-n} \Gamma (n+1,n)-1 \right)\\ &= n \end{aligned}$$

where $\Gamma$ is the incomplete Gamma function.

Answer 4

Since you asked " .. without using Gamma function .. " and if you just want to know whether the sum converges or not, then you were on the right track: just carry on with a BigO $$ \eqalign{ & \sum\limits_{k = 1}^n {\left( \matrix{ n \cr k \cr} \right)k!k{1 \over {n^{\,k} }}} = \sum\limits_{k = 1}^n {k{{n^{\,\underline {\,k} } } \over {n^{\,k} }}} = \sum\limits_{k = 1}^n {k{{n\left( {n - 1} \right)\quad \left( {n - k + 1} \right)} \over {n^{\,k} }}} = \cr & = \sum\limits_{k = 1}^n {k\left( 1 \right)\left( {1 - {1 \over n}} \right) \cdots \left( {1 - {{k - 1} \over n}} \right)} = 1 + \sum\limits_{k = 2}^n {k\left( {1 + O\left( {{1 \over n}} \right)} \right)} \quad \left| {\;2 \le n} \right. \cr} $$ to show that it diverges.