Let $\{a_n\}$ be a sequence of positive real numbers such that $\sum_{n=1}^\infty a_n$ is divergent. Which of the following series are convergent? Let $\{a_n\}$ be a sequence of positive real numbers such that
$\sum_{n=1}^\infty a_n$ is
divergent. Which of the following series are convergent?
a.$\sum_{n=1}^\infty \frac{a_n}{1+a_n}$
b.$\sum_{n=1}^\infty \frac{a_n}{1+n a_n}$
c.
$\sum_{n=1}^\infty \frac{a_n}{1+ n^2a_n}$
My Solution:-
(a)Taking $a_n=n$, then $\sum_{n=1}^\infty \frac{n}{1+n}$ diverges.
(b) Taking $a_n=n$,  $\sum_{n=1}^\infty \frac{n}{1+n^2}$ diverges using limit comparison test with $\sum_{n=1}^\infty \frac{1}{n}$
(c) $\frac{a_n}{1+n^2a_n}\leq \frac{a_n}{n^2a_n}=\frac{1}{n^2} $. Using comparison test. Series converges. I am not able to conclude for general case for (a) and (b)?

a. Use the $n$-th term test. $\lim_{n\to\infty} \frac{a_n}{a_n+1}=1$, regardless of whether $a_n$ increases without bound or is bounded. Because this is not equal to 0, it must diverge.
Better b. 
Use the limit comparison test with $\sum_{n=1}^{\infty} \frac{na_n}{a_n}=\sum_{n=1}^{\infty} n$. The limit comparison test, if applied correctly, gives a limit of 1, which satisfies the conditions. Because $\sum_{n=1}^{\infty} n$ diverges, the other series must diverge.
Keep in mind that the fact that the $a_n$ sum diverges was not used.
