Give an example of a convergent series $$\sum_{n=1}^{\infty}a_n$$ such that the series $$\sum_{n=1}^{\infty}a_{3n}$$ is divergent.. I cannot find find any kind of series... I am also be very thankful if you find a divergent series (changing the term convergent)...


2 Answers 2



  • $a_{3n} = \frac{1}{n+1}$
  • $a_{3n+1} = 0$
  • $a_{3n+2} = \frac{-1}{n+1}$

Then $\sum a_{3n}$ is the harmonic series and diverges.

Let $S_n=\sum_{k=0}^n a_k$, then, $\forall n\in \mathbb{N}, S_{3n} = \frac{1}{n+1}; S_{3n+1}=S_{3n+2}=0$ (try and prove it by induction if needed).

So, $\forall n \in \mathbb{N},0\leq S_n \leq \frac{1}{n+1}$ and by the squeeze theorem, since $\lim_{n\to \infty} \frac{1}{n+1}=0$, the series converges to $0$.

On the other hand, let:

  • $a_{3n} = \frac{1}{n^2}$
  • $a_{3n+1} = 1$
  • $a_{3n+2} = 1$

Then $\sum_{k=1}^{3n} a_k \geq 2n$ diverges but $\sum a_{3n}$ converges as a Riemann series.

  • $\begingroup$ How to show $\sum_{k=1}^{3n} a_k\leq \frac{1}{n}$ converges? $\endgroup$ May 30, 2018 at 14:33
  • $\begingroup$ The aim to find $\sum_{k=1}^{3n} a_k$ is to find the series of partial sum $S_{3n}$? $\endgroup$ May 30, 2018 at 14:37
  • $\begingroup$ Thank you for asking! That is actually an inconsistency (it is still true but not quite complete). To show that the whole series converges we actually need to show that $\sum_{k=1}^n \leq \frac{3}{n}$ which you can prove just by looking at the value of the partial sum up to $n$ depending on the value of $n \mod 3$. Then, once you have proved the inequality, you may use the squeeze theorem. $\endgroup$ May 30, 2018 at 14:45
  • $\begingroup$ Thanks a lot, but is there any easy way to show that $\sum_{k=1}^n a_k \leq \frac{3}{n}$ $\endgroup$ May 30, 2018 at 14:52
  • $\begingroup$ @user1942348 Please see my edit. I hope you will find it rigorous enough. If not, please do let me know. $\endgroup$ May 30, 2018 at 15:16

I am just coping/pasting my answer to this from a duplicate question.


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

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 $\sum_{n=1}^{\infty} \frac{\chi(n)}{n} = -1 + \sum_{k=1}^{\infty} \left( \frac{1}{3k} - \frac{1}{3k+1}\right)$.

If we note that $\frac{1}{3k}-\frac{1}{3k+1} = \frac{1}{3k(3k+1)}$ then a simple application of the comparison test yields the convergence of $\sum a_n$.


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