Does the converse of $n$th term test for divergence hold?
I am new to this topic so I don't have much idea about the proof. Please help.
I only know that the series converge if limit of the $n$th term of that sequence approaches $0$ as $n$ approaches $\infty$.
Let consider a convergent series
$$\lim_{n\to \infty} S_n=\lim_{n\to \infty}\sum_{k=1}^n a_k = L \in \mathbb R$$
then
$$\lim_{n\to \infty} \left(S_n-S_{n-1}\right)=\lim_{n\to \infty}\left( \sum_{k=1}^na_k-\sum_{k=1}^{n-1}a_k\right)=\lim_{n\to \infty} a_n =L-L=0$$
therefore we can conclude that
$$\sum_{k=1}^\infty a_k=L \in\mathbb R \implies a_n \to 0$$
Therefore $a_n \to 0$ represents a necessary condition for the convergence of the series.
To see that the converse implication doesn't hold let consider the counter example $a_n=\frac1n$ which leads to the divergent harmonic series.
Therefore $a_n \to 0$ doesn't represent a sufficient condition for the convergence of the series.