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Let $S=\lim_{n\to \infty} \sum_{k=0}^{n-1} \left(\frac{n-k}{n}\right)^n,$ then $S=\lim_{n\to \infty}\sum_{k=0}^{n-1} \left(1-k/n\right)^n=\lim_{n\to \infty}\sum_{k=0}^{n-1} \exp[n \ln (1-k/n)]$. Using $\ln(1-x)\sim -x$, when $x$ is very small, we get $$S=\sum_{k=0}^{\infty} e^{-k}=\frac{1}{1-1/e}=\frac{e}{e-1}.$$

As $k^2/(2n)$ has been ignored here: What could be other (better) method?

Z Ahmed
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2 Answers2

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Perhaps not better, but also rigorous. Fix $M \in \mathbb N$, and let $N_M \geq M$ be a positive integer large enough that $$\left(1-\frac{k}{N_M}\right)^{N_M} > e^{-k} -2^{-M}$$ for each $0\leq k\leq M$. Then it holds $$\frac{1-e^{-(M+1)}}{1-e^{-1}} - M2^{-M} \leq \sum_{k=0}^M \left(1-\frac{k}{N_M}\right)^{N_M} \leq \sum_{k=0}^{N_M} \left(1-\frac{k}{N_M}\right)^{N_M}.$$ Since $N_M \to \infty$ as $M \to \infty$, it follows $$\lim_{n\to \infty} n^{-n} \sum_{k=0}^n(n-k)^n \geq \frac{1}{1-e^{-1}}.$$ On the other hand, since $$\left(1-\frac{k}{N_M}\right)^{N_M} \leq e^{-k},$$ it holds $$\lim_{n\to \infty} n^{-n} \sum_{k=0}^n(n-k)^n \leq \frac{1}{1-e^{-1}},$$ completing the proof.

BBBBBB
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Let $a_n$ be the sequence $a_n(k) = (1-{k \over n})^n$ for $k = 0,...,n-1$ and zero subsequently, and $a$ be the sequence $a(k) = e^{-k}$. Note that $0 \le a_n(k) \le a(k)$ for all $k$, and $\lim_n a_k(n) = a_k$. (All sequences concerned are absolutely summable.)

Note that $| \sum_k a(k) - \sum_k a_n(k)| \le \sum_k |a(k)-a_n(k)|$, so if we can show that the latter term has limit zero then we can conclude that $\lim_n \sum_k a_n(k) = \sum_k a(k)$.

Let $\epsilon>0$ and choose $N$ such that $\sum_{k >N} |a(k)| < \epsilon$. Now choose $N' \ge N$ such that for $n \ge N'$ we have $\sum_{k \le N} |a(k)-a_n(k)| < \epsilon$ (note the summation is over the 'fixed' $N$ not $N'$ here). Then for $n \ge N'$ we have \begin{eqnarray} \sum_k |a(k)-a_n(k)| &\le & \sum_{k\le N} |a(k)-a_n(k)| + \sum_{k> N} |a(k)-a_n(k)| \\ & \le & \sum_{k\le N} |a(k)-a_n(k)| + 2\sum_{k> N} |a(k)| \\ &<& 3 \epsilon \end{eqnarray}

In particular $\lim_n \sum_k a_n(k) = \sum_k a_k$ and in this case the latter is the sum of a geometric series hence equal to ${1 \over 1-{1 \over e}}$.

This is essentially the dominated convergence theorem applied to sequences.

copper.hat
  • 172,524