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)?

If $\sum_{n=1}^{\infty} a_n$ be a divergent series of positive real numbers prove that the
series $\sum_{n=1}^{\infty}\frac{a_n}{1+a_n}$ is divergent.
Proof:
Let $S_n = a_1 +a_2 + ... +a_n$ .
Since the series $\sum_{n=1}^{\infty} a_n$ is a divergent series of positive real numbers, the
sequence $\{S_n\}$ is a monotone increasing sequence and $\lim S_n = \infty.$
Therefore for every natural number $n$ we can choose a natural number
$p$ such that $S_{n+p} > 1 + 2S_{n}$ . Now, $\frac{a_{n+1}}{1+a_{n+1}}+\frac{a_{n+2}}{1+a_{n+2}}+....+\frac{a_{n+p}}{1+a_{n+p}} >\frac{a_{n+1}}{1+S_{n+1}}+\frac{a_{n+2}}{1+S_{n+2}}+....+\frac{a_{n+p}}{1+S_{n+p}}$ (since  $S_{n+p}\ge S_{n+1}\ge a_{n+1}$....$S_{n+p}\ge a_{n+p}$)
Now $\frac{a_{n+1}}{1+S_{n+1}}+\frac{a_{n+2}}{1+S_{n+2}}+....+\frac{a_{n+p}}{1+S_{n+p}}= \frac{S_{n+p}-S_n}{1+S_{n+p}}>\frac{\frac{1}{2}(1+S_{n+p})}{1+S_{n+p}}=\frac{1}{2}$
This shows that Cauchy's principle of convergence is not satisfied by
the series $\sum_{n=1}^{\infty}\frac{a_n}{1+a_n}$ Hence the series is divergent.
