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What is $$ \lim_{n \to \infty} \sum_{k=0}^n \frac{{n\choose k}}{n^k(k+3)}\ ?$$ I know the way by integration and that the answer is $e-2$ but I am more interested in use of sandwich theorem which provides a maxima or a closed form to it. Expansions may also be useful.

Robert Z
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Archis Welankar
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  • Also that's not a sequence. –  Sep 02 '16 at 12:36
  • I fixed the limit display but it's unclear to me what the lower summation bound is supposed to be. $k=1$ maybe? Please update accordingly. –  Sep 02 '16 at 12:36
  • its n to infinity and sum is from 0 to n – Archis Welankar Sep 02 '16 at 12:39
  • For a closed form, you may remark that if $S_n$ is your sum, you have $S_n=\int_0^1 x^2(1+x/n)^ndx$, and putting $x=nu$ you get $S_n=n^3\int_0^{1/n}u^2(1+u)^n du$. Writing $u^2=(1+u)^2-2(1+u)+1$ and integrating gives you a closed form. But we have integrated... – Kelenner Sep 02 '16 at 13:42
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    For an upper bound, use the fact that $${n\choose k}\leqslant\frac{n^k}{k!}$$ hence the $n$th sum $S_n$ is such that $$S_n\leqslant\sum_{k=0}^n\frac1{k!(k+3)}<\sum_{k=0}^\infty\frac1{k!(k+3)}$$ For a lower bound, fix some $N$ and note that, for every $n\geqslant N$, $$S_n\geqslant\sum_{k=0}^N\frac{{n\choose k}}{n^k(k+3)}$$ Now, the RHS has a finite number of terms whose limit is clear, hence $$\liminf S_n\geqslant\sum_{k=0}^N\frac1{k!(k+3)}$$ The lower bound holds for every $N$, qed. – Did Sep 02 '16 at 16:28

2 Answers2

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For $0\leq k\leq n$, $$1-\frac{\binom{k}{2}}{n}\leq \prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)=\frac{k!}{n^k}\binom{n}{k}\leq 1$$ where the empty product is $1$. Therefore $$\sum_{k=0}^n \frac{1}{k!(k+3)}-\frac{1}{n}\sum_{k=0}^n \frac{\binom{k}{2}}{k!(k+3)}\leq \sum_{k=0}^n \frac{{n\choose k}}{n^k(k+3)}\leq \sum_{k=0}^n \frac{1}{k!(k+3)} \tag{1}$$ Then by noting that $\sum_{k=0}^{\infty} \frac{\binom{k}{2}}{k!(k+3)}$ converges, it follows that the sequence $\sum_{k=0}^n \frac{{n\choose k}}{n^k(k+3)}$ is bounded and, by applying the Squeeze Theorem in (1), we have $$\lim_{n \to \infty} \sum_{k=0}^n \frac{{n\choose k}}{n^k(k+3)}=\sum_{k=0}^{\infty}\frac{1}{k!(k+3)}=e-2.$$

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

Besides the 'sandwich' answer of $\texttt{@RobertZ}$, the following answer provides an alternative approach:

\begin{align} \lim_{n \to \infty}\sum_{k = 0}^{n}{{n \choose k} \over n^{k}\pars{k + 3}} & = \lim_{n \to \infty}\sum_{k = 0}^{n}{n \choose k} \pars{1 \over n}^{k}\int_{0}^{1}t^{k + 2}\,\dd t = \lim_{n \to \infty}\int_{0}^{1}t^{2}\sum_{k = 0}^{n}{n \choose k} \pars{t \over n}^{k}\,\dd t \\[5mm] & = \lim_{n \to \infty}\int_{0}^{1}t^{2}\pars{1 + {t \over n}}^{n}\,\dd t = \lim_{n \to \infty}\bracks{n^{3}\int_{0}^{1/n}t^{2}\pars{1 + t}^{n}\,\dd t} \end{align}

The last integral can be evaluated by succesive integration by parts which decreases the $\ds{t^{2}\mbox{-exponent}}$ to $\ds{0}$.

Namely, \begin{align} &\lim_{n \to \infty}\sum_{k = 0}^{n}{{n \choose k} \over n^{k}\pars{k + 3}} \\[5mm] = &\ \lim_{n \to \infty}\braces{n^{3}\bracks{-2n^{3} + \pars{1 + 1/n}^{n}\pars{1 + n}\pars{2 + n + n^{2}} \over n^{3}\pars{1 + n}\pars{2 + n}\pars{3 + n}}} = \bbx{\expo{} - 2} \end{align}

Felix Marin
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