To what value does this summation converge: $\sum_{r=o}^{n}{\frac{\binom{n-1}{r}r!}{n^{r+1}}}$

Question

I had been trying to solve (Probability of rolling a 1 before you roll two 2's, three 3's, etc) this problem for quite some while and I think I had found some way ahead but I cant seem to find the closed from for the summation that I ended up on. The summation is:

$$\frac{1}{n}\sum_{r=o}^{n-1}{\frac{\binom{n-1}{r}r!}{n^{r}}}$$

I also couldn't come up with worthy bounds for which the summation converges to a particular value(although it does converge as discussed in the original post) by the use of squeeze theorem.

Also WolframAlpha gives the sum as

$$\frac{1}{n}\sum_{r=o}^{n-1}{\frac{\binom{n-1}{r}r!}{n^{r}}}=\left({\dfrac{e}{n}}\right)^n\Gamma(n,n)$$

Edit:- I dont know how to handle the $\Gamma(n,n)$ and that is what I need help in which I forgot to mention and it also was the whole point of the post.

As written this is a finite sum which should always converge — TomGrubb, May 20 '17 at 02:19
@ThomasGrubb - Edited the post, I had not written it in a very neat manner. — user350331, May 20 '17 at 02:27
I suppose a typo; it should be $$\frac{1}{n}\sum_{r=o}^{n-1}{\frac{\binom{n-1}{r}r!}{n^{r}}}=e^n n^{-n} \Gamma (n,n)$$ — Claude Leibovici, May 20 '17 at 04:23
So, what is the problem ? The result for any $n$ is a constant, the value of which decreases smoothly when $n$ increases. — Claude Leibovici, May 20 '17 at 04:44
@ClaudeLeibovici - I mentioned in the edit what my problem was in the first place which I had totally forgot to mention. Also after reading your answer it seems that I still wouldn't be able to get what the summation would approach as $n\to \infty$ — user350331, May 20 '17 at 06:41
@user350331. Don't you think that, from my naswer, the wole thing behaves as $\sqrt{\frac{\pi }{2n}}$ for large values of $n$ ? — Claude Leibovici, May 20 '17 at 07:13
@ClaudeLeibovici - Then, wouldn't this mean that the probability asked in the post that I linked would approach zero — user350331, May 21 '17 at 03:51

score 2 · Answer 1 · answered May 20 '17 at 05:44

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{r = 0}^{n - 1}{{n - 1 \choose r}r! \over n^{r + 1}} & = \sum_{r = 0}^{n - 1}{n - 1 \choose r}\ \overbrace{\int_{0}^{\infty}t^{r}\expo{-nt}\,\dd t}^{\ds{r! \over n^{r + 1}}} = \int_{0}^{\infty}\expo{-nt}\sum_{r = 0}^{n - 1}{n - 1 \choose r}t^{r}\,\dd t \\[5mm] & = \int_{0}^{\infty}\expo{-nt}\pars{1 + t}^{n - 1}\,\dd t = \expo{n}\int_{1}^{\infty}t^{n - 1}\expo{-nt}\,\dd t = \expo{n}n^{-n}\ \overbrace{\int_{n}^{\infty}t^{n - 1}\expo{-t}\,\dd t}^{\ds{\Gamma\pars{n,n}}} \\[5mm] & = \bbx{\pars{\expo{} \over n}^{n}\,\Gamma\pars{n,n}} \end{align}

The two arguments $\ds{\Gamma}$ is the Incomplete Gamma Function.

Claude Leibovici · Accepted Answer · 2017-05-20T05:15:05.340

As I wrote in comments $$S_n=\frac{1}{n}\sum_{r=o}^{n-1}{\frac{\binom{n-1}{r}r!}{n^{r}}}=\left(\frac e n \right)^n \,\Gamma(n,n)$$ If you have a look here (formula $8.11.12$), you will find a series expansion for $\Gamma(n,n)$. Using it, you should get $$S_n=\sqrt{\frac{\pi }{2}} {\frac{1}{ n^{1/2}}}-\frac{1}{3 n}+ \sqrt{\frac{\pi }{2}} \frac{1}{12n^{3/2}}-\frac{4}{135 n^2}+ \sqrt{\frac{\pi }{2}} \frac{1}{288n^{5/2}}+\frac{8}{2835 n^3}+O\left(\frac{1}{n^{7/2}}\right)$$ which seems to be very accurate even from small values of $n$ as shown in the table below $$\left( \begin{array}{ccc} n & \text{exact} & \text{approximation} \\ 1 & 1.00000 & 1.00197 \\ 2 & 0.75000 & 0.75020 \\ 3 & 0.62963 & 0.62968 \\ 4 & 0.55469 & 0.55471 \\ 5 & 0.50208 & 0.50209 \\ 6 & 0.46245 & 0.46245 \\ 7 & 0.43116 & 0.43117 \\ 8 & 0.40563 & 0.40563 \\ 9 & 0.38426 & 0.38426 \\ 10 & 0.36602 & 0.36602 \end{array} \right)$$

To what value does this summation converge: $\sum_{r=o}^{n}{\frac{\binom{n-1}{r}r!}{n^{r+1}}}$

2 Answers2