Prove the following limit about $e$

Question

I need to show that: $$\lim_{n \to +\infty}\sum_{k=0}^{k=n-1}\left(\frac{n-k}{n}\right)^n = \frac{e}{e-1}$$I observed that taking the limit term by term gives the result, but of course this is not justified. This exercise assume no prior knowledge on uniform convergence of series. (I am new to series but I remember from calculus a result on limits that is justified in this case.) So I should be able to prove it following another path.

It also appears that a conversion to a definite integral is unfeasible. What do you suggest?

This has the form $\sum_{k=0}^{\infty} f_n(k)$ for $\lim_{n\rightarrow\infty}f_n(k)=e^{-k}$. — Michael, Feb 24 '20 at 13:49
You will certainly need $(\frac{n-k}{n})^n\rightarrow e^{-k}$. In fact for $k<n$ you can show $(\frac{n-k}{n})^n\leq e^{-k}$. So the main thing, first prove $\leq$ then prove $\geq$. Can you do one or more of those? — Michael, Feb 24 '20 at 13:59
Can you elaborate this as an answer? Because the last inequality you wrote isn't that obvious. — Peanut, Feb 24 '20 at 14:25
But using that last inequality, can you get somewhere? Prove $\leq e/(e-1)$? A useful inequality to know is $\log(1+x)\leq x$ for all $x>-1$. — Michael, Feb 24 '20 at 14:26
So then the other direction remains, just use $f_n(k)\geq 0$ for $k<n$ so $$\sum_{k=0}^{m-1} f_n(k) \leq \sum_{k=0}^{n-1} f_n(k)\quad \forall n>m$$ — Michael, Feb 24 '20 at 14:28
I can't show that there is some $x$ such that $xe^{-k} <= (1-k/n)^n$ and $\lim_{n \to \infty}x= 1$ where $x$ is a function of $n$ alone. This way I can use the squeeze theorem — Peanut, Feb 24 '20 at 15:10
As far as the other inequality is concerned, i.e. $(1-k/n)^n <= e^{-k}$ I used the log inequality you provided above. — Peanut, Feb 24 '20 at 15:12
Does this answer your question? How to evaluate $ \lim \limits_{n\to \infty} \sum \limits_ {k=1}^n \frac{k^n}{n^n}$?. Also asked earlier today: https://math.stackexchange.com/q/3558097/321264. — StubbornAtom, Feb 24 '20 at 15:13
@StubbornAtom yes I couldn't find it. But how did you find it? — Peanut, Feb 24 '20 at 15:14
Regarding your last comment "I can't show..." I agree that to get the lower bound this is good to show, that is why I suggested using $\sum_{k=0}^m f_n(k)\leq \sum_{k=0}^{n-1} f_n(k)$ for $n>m$. We fix an integer $m$. We know $f_n(k)\rightarrow e^{-k}$ so we can apply that to all of the $m$ terms in the sum $\sum_{k=0}^{n-1} f_n(k)$. Or, you can indeed say that for any $\epsilon>0$, there is an $n$ such that $f_n(k)\geq e^{-k} - \epsilon$ for all $k \in{0, ..., m-1}$. — Michael, Feb 24 '20 at 17:13

score 0 · Answer 1 · answered Feb 24 '20 at 15:22

COMMENT.- Except for some trick of which I do not believe, we must first find the form of $$S_n=\sum_{k=0}^{k=n-1}\left(\frac{n-k}{n}\right)^n$$ which is equal to $$S_n=1+\frac{1}{n^n}\left(1+2^n+3^n+\cdots+(n-1)^n\right)$$

So you have a problem that involves the sum of the $n-1$ first $n-\text {powers}$. Faulhaber's Formula gives $$S_n=1+\frac{1}{n^n}\left(\frac{(n-1)^{n+1}}{n+1}+\frac12((n-1)^n+\sum_{k=2}^n\frac{B_k}{k!}\frac{n!}{(n-k+1)!}n^{n-k+1}\right)$$ where $B_k$ is the Bernoulli number of index $k$ so you can simplify something the expresion because $B_{2k+1}=0$ and you have to consider only the $B_{2k}$. You can still move forward by examining theory about Bernoulli numbers.In particular the Bernoulli polynomials are defined by $$B_k(x)=\sum\binom nkB_{n-k}x^k$$ and the maybe important here is the generating function of these polynomials which is $$\frac{te^{tx}}{e^t-1}=\sum_{k=0}^{\infty}B_k(x)t^k$$

score 0 · Answer 2 · answered Feb 24 '20 at 17:19

To summarize my comments, we can define for each positive integer $n$: $$ f_n(k) = (1-k/n)^n \quad \forall k \in \{0, 1, ..., n-1\}$$ and we observe that for any positive integer $m$: $$ \sum_{k=0}^{m-1} f_n(k) \leq \sum_{k=0}^{n-1} f_n(k) \leq \sum_{k=0}^{n-1} e^{-k} \quad \forall n>m$$ So we can take limits as $n\rightarrow\infty$.

Prove the following limit about $e$

2 Answers2