If a series $\sum\lambda_n$ of positive terms is convergent, does the sequence $n\lambda_n$ converge to $0$? Let $\lambda_n>0, n\in\mathbb{N}$, with $\sum_n \lambda_n<+\infty$.
Can I conclude that $n\lambda_n\to 0$?
In this question and this question and their answers, it is shown that this is true if $\lambda_n$ are decreasing. What happens if $\lambda_n$ are not decreasing?
 A: No.
Define $\lambda_n$ by stating that $\lambda_{2^n}=2^{-n}$ and $\lambda_k=2^{-k}$ for other values of $k$.
Then $2^n\lambda_{2^n}=1$ so there is no convergence to $0$.
It is evident however that $\sum_n\lambda_n<\infty$.
A: Consider
$$
\lambda_n=\left\{\begin{array}{}
\frac1n&\text{if $n=k^2$ for some $k\in\mathbb{Z}$}\\
\frac1{n^2}&\text{if $n\ne k^2$ for any $k\in\mathbb{Z}$}\\
\end{array}\right.
$$
Then, when $n=k^2$,
$$
n\lambda_n=1
$$
yet
$$
\sum_{n=1}^\infty\lambda_n=2\zeta(2)-\zeta(4)
$$

However, if we have $\lambda_k\ge\lambda_{k+1}$, then
$$
\lim_{n\to\infty}n\lambda_n=0
$$
Suppose not. Then there is an $\epsilon\gt0$ so that for any $n$, there is an $N\ge n$ so that $N\lambda_N\ge\epsilon$. Then, because of the monotonicity, we have
$$
\begin{align}
\sum_{k=N/2}^{N}\lambda_k
&\ge\sum_{k=N/2}^{N}\frac\epsilon{N}\\
&\ge\frac\epsilon2
\end{align}
$$
and since we can choose $n$ as large as we want, there is a limitless set of sequences of terms whose sum is at least $\frac\epsilon2$. That is, we can choose $n_{j+1}=2N_j+2$ so that $N_{j+1}/2\ge n_{j+1}/2\gt N_j$, so that the intervals $[N_j/2,N_j]$ are disjoint and $\sum\limits_{k=N_j/2}^{N_j}\lambda_k\ge\frac\epsilon2$. Therefore,
$$
\sum_{k=1}^\infty\lambda_k=\infty
$$
Note: this latter argument is similar to this answer.
