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?

Question

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)?

Answer 1

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 $a_{n_k} \to \infty$

Answer 2

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.

Answer 3

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.