Give an example of a series $\sum a_n$ such that $\sum a_n$ is convergent but $\sum a_{3n}$ is divergent.

Question

Give an example of a series $\sum a_n$ such that $\sum a_n$ is convergent but $\sum a_{3n}$ is divergent.

Here an example by zhw. is given as $$\frac 1 2 + \frac 1 2 - \frac 2 2 + \frac 1 3 + \frac 1 3 − \frac 2 3 + \frac 1 4 + \frac 1 4 − \frac 2 4 + \frac 1 5 + \frac 1 5 − \frac 2 5+\cdots.$$ Give an example of a series such that $\sum a_n$ is convergent but $\sum a_{3n}$ is divergent

But I need a better result with functional expression.

Answer 1

Define

$$\chi(n) = \begin{cases} 1 & n \equiv 0 \mod(3) \\ -1 & n \equiv 1 \mod(3) \\ 0 & n \equiv 2 \mod(3) \end{cases}.$$

If we let $a_n = \frac{\chi(n)}{n}$ then $a_{3n} = \frac{1}{3n}$ and $\sum a_n$ converges while $\sum a_{3n}$ diverges.

Note: another way to write this is $\sum_{n=1}^{\infty} \frac{\chi(n)}{n} = -1 + \sum_{k=1}^{\infty} \left( \frac{1}{3k} - \frac{1}{3k+1}\right)$.

Edit: $\frac{1}{3k}-\frac{1}{3k+1} = \frac{1}{3k(3k+1)}.$ Use the comparison test.