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

  2. Give an example of a divergent series $\sum {b_n}$ such that the series $\sum {b_{3n}}$ is convergent.


  1. I am not sure if this is a valid forumla for a sequence : $ a_{3n-2} = \frac{1}{1+4(n-1)} ,a_{3n-1} = \frac{1}{3+4(n-1)}, a_{3n} = -\frac{1}{2n}$. This series converges to $\frac{3}{2} \log(2)$. But, $\sum {a_{3n}}$ diverges.

  2. We define $b_{3n-2}=1 , b_{3n-1}=1, b_{3n} = \frac{1}{n^2}$. The series diverges, but $\sum{b_{3n}}$ converges to $\frac{\pi^2}{6} $

The problem is, I am not sure if the this type of "formula" works [unlike the sequence defined by $1/n$ or something. Is this valid to define the sequence "term-by-term" (here, three different types of indices)?].

  • 7
    $\begingroup$ Yeah this is totally fine. $\endgroup$ – user113102 Apr 24 at 20:34

It is entirely fine to define a sequence term by term, and your examples are fine. In fact $\LaTeX$ even supports this with the following environment:

 [value 1] & [condition 1] \\
 [value 2] & [condition 2] \\

For example (right click to show underlying code): $$a_n= \begin{cases} 2&\text{ if }\ 3\text{ divides } n\\ -1&\text{ otherwise} \end{cases} \qquad\text{ and }\qquad b_n= \begin{cases} 0&\text{ if }\ 3\text{ divides } n\\ 1&\text{ otherwise} \end{cases}.$$

  • $\begingroup$ I will use it from now on. I was wondering how to do it. $\endgroup$ – Subhasis Biswas Apr 24 at 20:51

Term-by-term is fine. If you want an example of a series such that $\sum b_n$ diverges but $\sum b_{3n}$ diverges and $|b_{n+1}| \le |b_n|$ for each $i$, take $b_n = \frac{1}{n}$ iff $6 \not | n$ and $b_n = -\frac{1}{n}$ iff $6 | n$.


Other examples: For 1, consider the series

$$1+0+(-1) + \frac{1}{2} +0 +\frac{-1}{2} + \frac{1}{3} +0 +\frac{-1}{3}+\cdots $$

The series converges to $0,$ while $\sum a_{3n} = -\infty.$

For 2, consider the series

$$1 + 0 + 0 + 1 + 0 + 0 + 1 + 0 + 0 + \cdots$$


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