Does $a_k$ diverges I'm not sure, If I'm doing the convergence test correctly. From given,
$\sum_{k=1}^{\infty} b_{k}$ diverges and
$\sum_{k=1}^{\infty} \frac{a_{k}}{b_{k}}=2$
I have to check whether $\sum_{k=1}^{\infty} a_{k}$ diverges too?
My solution:
$\sum_{k=1}^{\infty} \frac{a_{k}}{b_{k}} = 2$
$\frac{\sum_{k=1}^{\infty} a_{k}}{\sum_{k=1}^{\infty} b_{k}} = 2$ Can i split a series like this? numerator and denominator seperately?
$\\ \sum_{k=1}^{\infty} a_{k} = 2 {\sum_{k=1}^{\infty} b_{k}} \implies \sum_{k=1}^{\infty} a_{k} \gt {\sum_{k=1}^{\infty} b_{k}}\\ $
By comparison test, 
WKT,
$\sum_{k=1}^{\infty} b_{k}$ diverges $\implies \sum_{k=1}^{\infty} a_{k}$ diverges too.
 A: You can't say anything :


*

*case 1 : take $b_k = n$, $a_k = \frac{\pi^2}{3} \frac{1}{n}$ : you don't have convergence of $\sum a_n$

*case 2 : take $b_k = 1$, $a_k = \frac{1}{2^{n-2}}$ : you have convergence of $\sum a_n$
Basically, as you can make $b_n$ diverge as fast as you want, $\frac{1}{b_n}$ can converge to $0$ as fast as you want, 
A: From the second series you get $\frac{a_n}{b_n} \to_n 0$, but this merely says that $a_n =o(b_n)$. For example, if $b_n=n^3$, then $a_n=O(n)$ or $a_n=O(\frac{1}{n^2})$ can both diverge and converge with $\sum_n \frac{a_n}{b_n}$ converging
A: Firstly no you cannot split a series like that, because $$a/b+c/d+e/f\neq \frac{a+c+e}{b+d+f}$$
Now the second comparison test says that if $$\lim_{n\to \infty}a_n/b_n=L\neq 0$$
Then $\sum_{k=1}^{\infty} a_k$ converges if and only if $\sum_{k=1}^{\infty} b_k$ converges. However in this case as $\sum_{k=1}^{\infty} \frac{a_{k}}{b_{k}} = 2$ we get that
$$\lim_{n\to \infty}a_n/b_n=0$$ and so the second comparison test is useless to us. The purpose of this question might be to actually reinforce the limitations of the comparison test for you.
