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?

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 $a_{n_k} \to \infty$

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Sequences And Series

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