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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)?

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  • $\begingroup$ Your solution is enough as it is: (a), (b) are not (surely) convergent, (c) is. $\endgroup$ – Quang Hoang Jun 5 '19 at 3:50
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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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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.

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  • $\begingroup$ what if $a_n = 1/n$? $\endgroup$ – Quang Hoang Jun 5 '19 at 3:49
  • $\begingroup$ I fixed that with the new solution $\endgroup$ – Math Jun 5 '19 at 3:55
  • $\begingroup$ The limit $\lim_{n \to \infty} \frac{a_n}{a_n + 1}$ need not be $1$. If $a_n \to 0$, then the limit is $0$. $\endgroup$ – Theo Bendit Jun 5 '19 at 4:01

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