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.
Best Answer
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.