Prove if $a_n$ is non-negative and $\sum a_n$ converges then $a_n\leq\frac{1}{n}$ for all $n\geq N$ for some some N

Question

I believe this is intuitively true but I cannot figure out how to prove it. I was thinking I could use a proof by contradiction. If there is no such $N$ then there are an infinite number of terms of $a_n$ that are greater than or equal to $\frac{1}{n}$. I can't prove that $a_n$ diverges, however. How do I go about proving this statement.

Answer 1

No, this statement is false.

We have the following counter-example

\begin{cases} \frac{1}{n^2} \qquad \text{if } \color{red}{\not \exists} k \in \Bbb N^* \text{ such that } n= k^4 \\ \frac{1}{\sqrt{n}} = \frac{1}{k^2} \qquad \text{if } \color{red}{\exists} k \in \Bbb N^* \text{ such that } n= k^4 \end{cases}

Hence, the sum of series $\sum a_n$ converges because

\begin{align} \sum_{n=1}^{+\infty} a_n & = \sum_{n\ne k^4}\frac{1}{n^2}+\sum_{n= k^4}\frac{1}{\sqrt{n}} \\ & = \sum_{n\ne k^4}\frac{1}{n^2}+\sum_{k \in \Bbb N^*}\frac{1}{k^2} < \sum_{n\in \Bbb N^*}\frac{1}{n^2}+\sum_{k \in \Bbb N^*}\frac{1}{k^2} < +\infty\\ \end{align}

and for all $N$, we have always $n = k^4$ with $k>N$ such that

$$a_n =\frac{1}{\sqrt{n}} > \frac{1}{n}$$

Answer 2

$a_n=\frac 2 n$ when $n$ is of the form $m^{2}$ for some $m$ and $0$ otherwise gives a counter-example.