How to prove that $\sum_{n=1}^{\infty}\frac{x}{n}\left(1+\frac{1}{n}-x\right)^n$ converges uniformly.

Question

Prove that the following functional series converges uniformly for any $x$ from $E$. $$ \sum_{n=1}^{\infty}u_n(x),\ \ \ u_n(x)=\frac{x}{n}\left(1+\frac{1}{n}-x\right)^n,\ \ x\in E=[0;1] $$

I tried to use Weierstrass M-test. So, I did the following: $$ |u_n(x)|\leqslant \frac{1}{n}\left(1+\frac{1}{n}\right)^n=a_n $$ However, $\sum_{n=1}^{\infty}a_n$ diverges. Thus, I have to find another solution. Perhaps I can find $v_n(x):|u_n(x)|\leqslant v_n(x)$ where $\sum_{n=1}^{\infty}v_n(x)$ converges uniformly. However, I have not succeeded so far.

Answer 1

Hint

If you look at the derivative of $u_n(x)$, you'll find that $\vert u_n(x) \vert$ is having a maximum at $x_n = \frac{1}{n}$. As you have $u_n(\frac{1}{n})=\frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges you get the result as a consequence of Weierstrass M-test.

Answer 2

It is not hard to prove that $u_n$ attains its maximum at $\frac1n$. Besides$$u_n\left(\frac1n\right)=\frac1{n^2}.$$Since the series $\sum_{n=1}^\infty\frac1{n^2}$ converges…

Answer 3

$u_n(x)$ can also be estimated using the inequality between the geometric and the arithmetic mean: $$ 0 \le n^2 u_n(x) = (nx) \left(1+\frac{1}{n}-x\right)^n \le \left( \frac{nx + n(1+\frac 1n - x)}{n+1} \right)^{n+1} \le 1 \, . $$