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

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