# Are there sequence $(u_n)\subset \mathbb R^+$ s.t. $\sum_{n=1}^\infty u_n<\infty$ but $nu_n\not\to 0$?

I have an exercise that ask me to prove that if $(u_n)$ is decreasing, $u_n\geq 0$ for all $n$ and $\sum_{n=1}^\infty u_n$ converge, then $nu_n\to 0$. I proved it, but for me this result in fact correct even if we remove $(u_n)$ decreasing. Because I have in mind that $1/n$ is the quickest speed of non convergence, i.e. if $\sum_{n}v_n$ converge, it will have $\alpha$ s.t. $v_n\leq \frac{1}{n^\alpha }<\frac{1}{n}$ for all $n$ for a certain $n$. But if we impose $u_n$ decreasing, maybe my imagination is wrong. So is there a positive sequence s.t. $\sum_{n=1}^\infty u_n$ converge but $nu_n\not\to 0$ ?

• Do you mean $1/n$ is the slowest speed of non-convergence? Commented Jul 30, 2018 at 19:41
• @packetpacket I think maybe he is referring to $n$, i.e. if $n$ would grow any faster then the series would converge. Some people think of the series about how fast the denominator is growing, but do not explicitly state that. Commented Jul 31, 2018 at 0:55

To answer the question in your title. For each positive integer $t$ set $u_{t^2} = \frac{1}{t^2}$. And for all nonsquare positive integers $n$ set $u_n = \frac{1}{n^3}$. Then

1. On the one hand $\sum_{j=1}^{\infty} u_j$ converges (as $\sum_{t=1}^{\infty} \frac{1}{t^2}$ converges as does $\sum_{n=1}^{\infty} \frac{1}{n^3}$).

2. On the other hand, $t^2u_{t^2} = 1$ for all positive integers $t$ so $nu_n$ does not converge to 0 as $n$ goes to infinity. i.e., No matter how large $n$ is, there is a positive integer $n'$ s.t. $n'u_{n'} =1$, namely $n'$ is any perfect square larger than $n$.

The $u_n$s are not nonincreasing though.

You can prove that if $a_n$ is decreasing and $na_n\not\to 0$, then $\sum a_n$ diverges.

Proof: If $na_n\not\to0$, then there is an $\epsilon>0$ and infinitely many indices $n(1)<n(2)<\dots$ for which $a_{n(i)}\ge \epsilon/n(i)$. Since $a_n$ is decreasing, we can group terms based on which of the intervals $[1,n(1)),[n(1),n(2)),\dots$ they lie in, and you get $$\sum_n a_n \ge \sum_i (n(i+1)-n(i))\cdot \frac{\epsilon}{n(i)}\stackrel{*}=-\sum_i n(i+1)\Big(\frac{\epsilon}{n(i+1)}-\frac{\epsilon}{n(i)}\Big)=\epsilon \Big(\sum_i \frac{n(i+1)}{n(i)}-1\Big)\tag{1}$$ where $\stackrel{*}=$ follows by summation by parts (check this!).

Now, let $r(i) = \frac{n(i+1)}{n(i)}$, with the convention $r(0)=n(1)$. It is a fairly standard result that $$\sum_i r(i)-1<\infty \iff \prod_i r(i)<\infty$$ which follows from $\log x\sim x-1$ as $x\to1$. However, since $\prod_{i=0}^k r(i)=n(k+1)\to\infty$, we must have that the final sum in $(1)$ diverges.