Finding $a_n\to 0$ such that $\sum a_n/n$ diverges

Question

Can someone please help me find a counter example for the following claim?

If $a_n \to 0$ then $ \sum_ 1 ^{\infty} \frac{a_n}{n}$ converges.

Answer 1

Take, for example, $$ a_n=\frac1{\log n} $$ for $n>1$. We can use the integral test for convergence to make sure that such series diverges.

Answer 2

We could also argue this way: For $k=1,2,\dots,$ define blocks of integers $B_k = \{2^k+1, 2^k+2,\dots , 2^{k+1}\}.$ Note that $B_k$ contains exactly $2^k$ integers. Define $a_n = 1/k, n \in B_k.$ Then $a_n \to 0$ and

$$\sum_{n\in B_k} a_n/n = (1/k)\sum_{n\in B_k} 1/n \ge (1/k) [2^k\cdot (1/2^{k+1})] = 1/(2k).$$

Thus $\sum_n a_n/n = \sum_k\sum_{n\in B_k} a_n/n \ge \sum_k 1/(2k) = \infty.$