real analysis - 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)?].

Answer

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):

$$a_n= \begin{cases} 2&\text{ if }\ 3\text{ divides } n\\ -1&\text{ otherwise} \end{cases} \qquad\text{ and }\qquad b_n= \begin{cases} 0&\text{ if }\ 3\text{ divides } n\\ 1&\text{ otherwise} \end{cases}.$$