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)?
Best Answer
For (b) the series could diverge as you showed or converge as with
$$a_n = \begin{cases} 1, & n = m^2 \\ \frac{1}{n^2}, & \text{otherwise}\end{cases}$$
since
$$\sum_{n= 1}^N \frac{a_n}{1+na_n} = \sum_{n \neq m^2} \frac{a_n}{1+na_n} + \sum_{n = m^2} \frac{a_n}{1+na_n} \\ \leqslant \sum_{n= 1}^N \frac{1}{n + n^2}+ \sum_{n= 1}^N \frac{1}{1+n^2}$$
For (a) the series always diverges.
Consider cases where $a_n$ is bounded and unbounded. If $a_n < B$ then $a_n/(1 + a_n) > a_n/(1+B)$ and we have divergence by the comparison test.
Try to examine the second case where $a_n$ is unbounded yourself. Hint: There is a subsequence $a_{n_k} \to \infty$