If a series $\sum\lambda_n$ of positive terms is convergent, does the sequence $n\lambda_n$ converge to $0$?
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 $\sum_n\lambda_n<\infty$.
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.