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

Question

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.

How about $\sum_{k=1}^\infty\frac{\cos(2k\pi/3)}{k}$? – Dejan Govc May 30 '18 at 14:43 — Dejan Govc, May 30 '18 at 14:43
@DejanGovc Elaborate it please – user1942348 May 30 '18 at 14:44 — user1942348, May 30 '18 at 14:44

user328442 · Accepted Answer · 2018-05-30T15:09:28.857

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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.

edited May 30 '18 at 15:09

answered May 30 '18 at 14:45

user328442

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@user3228442 Nice, but how to say that ∑$a_n$ converges? – user1942348 May 30 '18 at 14:48
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@user328442 It's $\text{\begin{cases} then type whatever your cases are (with double backslashes to make new lines),}\\text{ then \end{cases}} $when you're done. – Dave May 30 '18 at 15:04
@Dave I’ll make the edit then. Thanks! – user328442 May 30 '18 at 15:05

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