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

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?

• The answer to $a$ is correct. Good job Commented Sep 17, 2016 at 19:25
• for those exercices, in general you have to use the partial summation $\sum_{n=1}^N a_n u_n = v(N)a_N +\sum_{n=0}^{N -1} v(n)(a_n-a_{n+1})$ where $v(N) = \sum_{n=1}^N u_n$. since $\sum_n u_n$ converges, $|v(n)| < C$, and with $a_n = u_n n$, you have $\sum_{n=0}^{N -1} v(n)(a_n-a_{n+1}) = \sum_{n=0}^{N -1} v(n)(u_n n-u_{n+1} n) < \sum_{n=0}^{N -1} C|u_n|$. So it reduces to $v(N)u_N N \to 0$, i.e. $u_n = o(1/n)$, which is the case since otherwise ($u_n$ being decreasing) $\sum_n u_n$ wouldn't converge Commented Sep 17, 2016 at 20:03
• and since $u_n = o(1/n)$, you have $n = o(1/u_n)$ and $\frac{u_n}{1-n u_n} =\frac{1}{ \frac{1}{u_n}-n} \sim u_n$ so yes $\sum_n \frac{u_n}{1-n u_n}$ converges (for b) note how $u_n = o(1/n)$ was important) Commented Sep 17, 2016 at 20:05
• For (b) suppose $a_{(4^n)}=1/2^n$ and $a_m=1/2^m$ when $m$ is not a power of $4.$ Then $\sum_{j\in \Bbb N} a_j<\sum_{n \in \Bbb N}1/2^n+\sum_{m\in \Bbb N}1/2^m=2$. But $4^n(a_{(4^n)})^2=1.$ Commented Jan 29, 2019 at 6:27
• The idea in my previous comment is that we can have $a_j=1/\sqrt j$ for a "sparse" but infinite collection of $j$'s and still have $\sum_na_n<\infty.$ Commented Jan 29, 2019 at 6:34

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.