I think I have proved a theorem but I am not sure if my solution is right.
Theorem: A necessary condition for the convergence of a series $S =\sum a_n$ is that $a_n\to 0$ as $n\to \infty$.
My proof:
Suppose the contrary that $a_n \to L\neq 0$ for as $n$ goes to infinity. From the definition of the limit we can say that
$$|a_n-L|<\max\{|a_i-L|, i=1,2,3,…,K\}, $$
where $K$ a number such as for $n > K\Rightarrow |a_n-L|<1$, let's call this number $M$. Then
$$L-M < a_n < L+M. $$
Let $S_1$ be the series of the sum of $L-M$ and $S_2$ be the series of the sum of $L+M$, then $S_1 < S < S_2$, but $S_1$ and $S_2$ diverge so $S$ must diverge too, which is false so $a_n$ must go to 0 as $n$ goes to infinity.
I do not know if this is correct.
Best Answer
You cannot assume that $\{a_n\}$ is convergent. The contrary of $a_n\to0$ is not that $a_n$ converges to another number, but rather that the sequence does not approach zero. Concretely, it means that there exists $\varepsilon>0$ such that for every $N$ there exist $n>N$ with $|a_n|\geq\varepsilon$. This allows you to choose an infinity of $a_n$ (propertly, a subsequence) that is bounded away from zero, and this will contradict the convergence of the series.
A more direct proof is as follows: since $\sum_na_n$ is convergent, then for each $\varepsilon>0$ there exists $N$ such that for any $N<k<m$ we have $\left|\sum_{k\leq n\leq m}a_n\right|<\varepsilon$. So we can choose, for each $\varepsilon >0$, an $N$ such that for all $k>N$ we have $|a_k|=\left|\sum_{k\leq n < k+1}a_n\right|<\varepsilon$. That is, $a_n\to0$.