# 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)?

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

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 $$a_{n_k} \to \infty$$

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.

• what if $a_n = 1/n$? – Quang Hoang Jun 5 '19 at 3:49
• I fixed that with the new solution – Math Jun 5 '19 at 3:55
• 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 $0$. – Theo Bendit Jun 5 '19 at 4:01