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 $nu_n\not\to 0$ ?
Are there sequence $(u_n)\subset \mathbb R^+$ s.t. $\sum_{n=1}^\infty u_n<\infty $ but $nu_n\not\to 0$
real-analysissequences-and-series
Best Answer
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
On the one hand $\sum_{j=1}^{\infty} u_j$ converges (as $\sum_{t=1}^{\infty} \frac{1}{t^2}$ converges as does $\sum_{n=1}^{\infty} \frac{1}{n^3}$).
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.