# Series convergence $\sum\limits_{k = 1}^{+\infty} k a_k^2$

I found a nice problem about series convergence. And would like to share it.

Suppose series $\sum\limits_{k = 1}^{+\infty} a_k$ converges absolutely.

1) Found an example when $\sum\limits_{k = 1}^{+\infty} k a_k^2$ diverges.

2) Suppose that $a_k$ is a non increasing sequence. Is it true that $\sum\limits_{k = 1}^{+\infty} k a_k^2$ converges?

3) Is it true that if $a_k$ is a non increasing then $\sum\limits_{k = 1}^{+\infty}k^2 a_k^2$ converges?

• For (1) let $a_k=0$ when $k$ is not the $4$th power of a natural number and $a_{n^4}=1/n^2$ for $n\in \mathbb N.$ Then $\sum_{k=1}^{\infty}a_k=\pi^2 /6$. But $n^4 (a_{n^4})^2=1$ for $n\in \mathbb N$. – DanielWainfleet Jun 14 '17 at 0:55

I would like to know simplier solution on second case but I will try to tell my solutions under spoilers.

1)

1) $a_{k^8} = 1/k^2$ and $a_t = 0$ if $t \neq k^8$. Then obviously series converges but $\sum\limits_{k = 1}^{+\infty} k^6 = +\infty$.

3)

Take $a_k = \frac{1}{k\log^2 k}$ and $k \geq 2$ (it does not matter at all). And series is convergent (and sequence is non-increasing) but it is obviously that $\sum\limits_{k = 2}^{+\infty} \frac{1}{\log^4 k} = +\infty$.

2) There are two cases. Each is under spoiler for challenging :)

First case

First assume that the number of indexes $k$ such that $|a_k| \geq 1/k$ is finite. Then $\exists N \ \forall n > N \implies |a_n| < 1/n$. But then $n a_n^2 < |a_n|$ and series $\sum\limits_{k = 1}^{+\infty} k a_k^2$ converges.

Second case

Let's find a subsequence of indexes such that $|a_{k_n}| \geq 1/k_n$. Now we will work with non-negative numbers and have a contradiction with absolute convergence. Then we will change our subsequence in a way that $\frac{k_{n + 1}}{k_n} > 2$ (we can do this due to infinite large indexes property). Then let's replace all $a_k$ by $a_{k_n}$ where $k_n \geq k$ and $k_n - k$ is minimal through all $n$. Then we replaced all elemets by less or equal ones. Clearly, series of new $a_k$ diverges (on each segment $(k_{n - 1}, k_n]$ we sum at least $1/k_n$ at least $k_{n} - k_{n - 1} \geq k_{n - 1}$ times. So sum is at least 1/2 on each segment).

• The second case of 2) cannot occur because if $(a_k)_k$ is monotonic and $\sum_ka_k$ converges then $\lim_{k\to \infty}ka_k=0.$ – DanielWainfleet Jun 14 '17 at 1:02

I suggest the following solution:

1) For the first part, I like your solution of setting $a_{k^8}=1/k^2$ and $a_t=0$ if $t≠k^8$, I don't think it gets much simpler.

2) Due to this question, we know that under the stated assumptions $\lim_{k\to\infty} ka_k = 0$. In particular, there is a $n_0 \in \mathbb{N}$ such that $ka_k = |ka_k| < 1$ for all $k > n_0$, so we have

$$\sum_{k=n_0}^{\infty} ka_{k}^2 \leq \sum_{k=n_0}^{\infty} a_k < \infty.$$ 3) $a_k = k^{-3/2}$ is a counterexample.