# 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.

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