Give an example of a convergent series $\sum {a_n}$ such that the series $\sum {a_{3n}}$ is divergent.
Give an example of a divergent series $\sum {b_n}$ such that the series $\sum {b_{3n}}$ is convergent.
Attempt:
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.
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)?].