Understanding $(\frac{1}{n})^n+(\frac{2}{n})^n+…+(\frac{n}{n})^n$ sum

Problem: $$(\frac{1}{n})^n+(\frac{2}{n})^n+...+(\frac{n}{n})^n$$ converges to $\frac{e}{(e-1)}$ as $n\to\infty$

Author's solution:

Let $d_n$ be any monotone increasing sequence of positivw integers diverging to $\infty$ and satisfying $d_n<n$ for $n>1$. We divide the sum into parts as follows:$a_n=(\frac{1}{n})^n+(\frac{2}{n})^n+...+(\frac{n-1-d_n}{n})^n\\b_n=(\frac{n-d_n}{n})^n+...+(\frac{n}{n})^n$

First the sum $a_n$ is roughly estimated above by:

$\frac{1}{n^n}\int\limits_{0}^{n-d_n}x^n dx=\frac{n-d_n}{(n+1)n^n}<(1-\frac{d_n}{n})^n$

Now using an inequality $\log(1-x)+x<0$ valid for $0<x<1$ we obtain:

$0<a_n<e^{n\log(1-\frac{d_n}{n})}<e^{-d_n}$

which converges to $0\to\infty$. Next by Taylor's formula $\log(1-x)$ we can take a positive constant $c_1$ such that $|\log(1-x)+x|\leqslant c_1x^2$ holds for $|x|<\frac{1}{2}$. This for any integer $n$ satisfying $\frac{d_n}{n}\leqslant \frac{1}{2}$ we get:

$|\log(1-\frac{k}{n})+k|\leqslant \frac{c_1k^2}{n}\leqslant\frac{c_1d_n^2}{n}$

for $0\leqslant k\leqslant d_n$. Suppose further that $\frac{d^2_n}{n}$ converges to $0$ as $n\to\infty$, For example $d_n=[n^{\frac{1}{3}}]$ satisfies all the conditions imposed above. Next take positive constant $c_2$ satisfying $|e^x-1|\leqslant c_2|x|$

for any |x|\leqslant 1. Since $\frac{c_1d_n^2}{n}\leqslant 1$ for all sufficiently large $n$, we have:

$|e^k(1-\frac{k}{n})^n-1|=|e^{n\log(1-k/n)+k}-1|\leqslant \frac{c_1c_2d_n^2}{n}$

Dividing both sides by $e^k$ and summing from $k=0$ to $d_n$, we get:

$\sum_\limits{k=0}^{d_n}|(1-\frac{k}{n})^n-e^k|\leqslant \frac{c_1c_2d^2_n}{n}\sum_\limits{k=0}^{d_n}e^{-k}$.

$|b_n-\sum_\limits{k=0}^{d_n}e^{-k}|<\frac{ec_1c_2d^2_n}{(e-1)n}$,

which implies that:

$|b_n-\frac{e}{e-1}|\leqslant \frac{e}{e-1}(\frac{c_1c_2d^2_n}{n}+e^{-d_n})$.

Therefore $a_n+b_n$ converges to $e/e-1$ as $n\to\infty\blacksquare$.

I am trying to solve some difficult real analasys problems. Since I am starting I am trying to develop my skills by glimpsing to the resolution in the book. However this problem seems to me as very difficult and I am stuck at its proof.I have written all the proof despite the fact I am stuck at the beginning.

Questions:

1) Why is it defined $d_n$ in the proof as a diverging sequence? Why part the sequence into two?

2) How did the author proceed to get this $\frac{n-d_n}{(n+1)n^n}<(1-\frac{d_n}{n})^n$ inequality?

3) I have tried to prove $|\log(1-x)+x|\leqslant c_1x^2$ using Taylor expansion but I was not able to do it. How do I prove it?

• It looks like there are some questions about this sum which were already asked one this website. Namely How to prove that: $\lim_{n\to\infty} \frac{1^n+2^n+\cdots+n^n}{n^n} = \frac{e}{e-1}$ and How to evaluate $\lim \limits_{n\to \infty} \sum \limits_ {k=1}^n \frac{k^n}{n^n}$?. Maybe this could help you somehow. – mrtaurho Aug 7 '18 at 21:38
• (1) The original solution tries to hide the idea of the problem already from the start, just skip this proof of an enigmatic style. The idea is that at a later point, the solution will choose a "special" $d_n$, but at the start we only need $d_n\to\infty$. (2) There is a missing $(n+1)$-power of the numerator. Then use $n-d_n<n+1$. (3) Use the Taylor polynomial of order one,. – dan_fulea Aug 7 '18 at 22:20
• @dan_fulea Thanks for your help! I am still stuck at 3).$\log(x-1)=-x-O(x^2)$, then $\log(x-1)+x=-x-O(x^2)+x\implies \log(x-1)+x=O(x^2)\implies|\log(x-1)+x|=O(x^2)\implies |\log(x-1)+x|\leqslant c_1x^2$. However the author points out that the latter identity only holds for $|x|\leqslant\frac{1}{2}$. This later point did not show in my derivation. What am I doing wrong? Thanks in advance! – Pedro Gomes Aug 9 '18 at 21:25