$\sum_{n}^{\infty}u_n$ converges $\Rightarrow $ $\sum_{n}^{\infty }nu_n^2$ and $\sum_{n}^{\infty}\frac{u_n}{1-nu_n}$ coverge?

Question

Let $(u_n)_{n\geq 1}$ be a sequence with $u_n\in \mathbb{R}_+$, such that $\sum_{n}^{\infty}u_n$ converges.

a)Suppose that $(u_n)_{n\geq 1}$ decreases:

1,Prove that $nu_n\rightarrow 0$ as $n\rightarrow \infty$.

2,Do the series $\sum_{n}^{\infty }nu_n^2$ and $\sum_{n}^{\infty}\frac{u_n}{1-nu_n}$ coverge?

b) Examine 2, when not supposing that $(u_n)_{n\geq 1}$ decreases.

I'm not sure about my solution for a)

We have $0\leq \frac{n}{2}u_n\leq \sum_{k=\left \lfloor \frac{n}{2} \right \rfloor}^{n}u_k\rightarrow 0$, $\Rightarrow nu_n\rightarrow 0 $ as $n\rightarrow \infty$.

$\exists M\in\mathbb{R}_+:\forall n\in\mathbb{N}^*, nu_n\leq M$ $\Rightarrow 0\leq nu_n^2\leq Mu_n$, hence: $\sum_{n}^{\infty }nu_n^2$ converges.

$\frac{u_n}{1-nu_n}\sim u_n \ \ as \ \ n\rightarrow \infty$, hence: $\sum_{n}^{\infty}\frac{u_n}{1-nu_n}$ coverges.

I can't solve b). Could you give me some hints?

Answer 1

The $\sum_n^\infty u_n= \sum_n^\infty nu_n \frac{1}{n}$, if $n u_n \to k>0$ then the sum is like $\sum_n^\infty \frac{k}{n}= \infty$, and i didn't use the decrease argument. If $\sum_{n=p}^\infty nu_n^2\le \sum_{n=p}^\infty u_n< \infty$ if $p$ is large enough, and if $p$ is such that $1 - nu_n> 1- \epsilon$ for all $ n\ge p$ then $\sum_{n=p}^\infty \frac{u_n}{1- nu_n}\le \sum_{n=p}^\infty \frac{u_n}{1-\epsilon}$, hence 2) is still valid without the decrease. It is not valid if we eliminate the $u_n \in \mathbb{R}_+$ hiypothesis.