A series such that $\sum {a_n}$ converges, but $\sum {a_{3n}}$ diverges.

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.

Attempt:

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)?].

• Yeah this is totally fine. – 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:

a_n=
$$\begin{cases} [value 1] & [condition 1] \\ [value 2] & [condition 2] \\ ... \end{cases}$$


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

• I will use it from now on. I was wondering how to do it. – 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

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