Help me to solve the limit $\lim_{n \to \infty} \Big[\Big( \frac 1 n\Big)^n+\Big(\frac 2 n\Big)^n + \dots +\Big(\frac n n\Big)^n\Big]=\dots $

Question

Help me with this limit to infinity question (https://i.stack.imgur.com/Jo52z.jpg)

$$\lim_{n \to \infty} \Big[\Big( \frac 1 n\Big)^n+\Big(\frac 2 n\Big)^n + \dots +\Big(\frac n n\Big)^n\Big]=\dots $$

Answer 1

Boy good thing I put in that false answer ! This way I can still solve the problem even after its closes. LOOPHOLE !

Ok so the limit is

$$x=\lim\limits_{n\to \infty} \sum\limits_{k=0}^n \left(1-\frac{k}{n}\right)^n$$ now if we just take the first $k$ terms as a lower bound be get

$$1+\left(1-\frac{1}{n}\right)^n+\cdots +\left(1-\frac{k}{n}\right)^n\leq x$$

And taking the limit with $k$ fixed gives

$$1+\frac{1}{e}+\cdots +\frac{1}{e^k}\leq x$$

On the other hand $y_n=\left(1-\frac{k}{n}\right)^n$ is an increasing sequence with $k$ fixed (I checked it with Bernoulli's inequality). So

$$\left(1-\frac{k}{n}\right)^n\leq \frac{1}{e^k}$$ and thus

$$\sum\limits_{k=0}^{\infty}\frac{1}{e^k}$$ is an upper bound.

So the limit is $$\frac{e}{e-1}$$

Answer 2

If $k$ is small compared with $n$, $(\frac{n-k}{n})^n =(1-k/n)^n \approx e^{-k} $.

So the sum of the last $k$ terms is about $\sum_{j=0}^{k-1} e^{-j} =\dfrac{1-e^{-k}}{1-e^{-1}} \approx \dfrac1{1-1/e} $ for large $k$.

Since $-\ln(1-x) > x$, $1-x < e^{-x} $ so $1-k/n < e^{-k/n}$ or $(1-k/n)^n < e^{-k}$.

Therefore the sum of the terms after the $k$-th is $\sum_{j=k}^{n}(1-j/n)^n \lt \sum_{j=k}^{n-1} e^{-j} =\dfrac{e^{-k}-e^{-n}}{1-e^{-1}} \lt \dfrac{e^{-k}}{1-1/e} \to 0 $ for large $k$.

Therefore, and I think this can be made rigorous, the sum is $\dfrac1{1-1/e} \approx 1.5819767068693265$.

Wolfy seems to agree.