# $\sum_{n=1}^{\infty}a_{n}$ diverges but $\sum_{n=1}^{\infty}\frac{a_{n}}{1+a_{n}^{2}}$ sometimes converges and sometime diverges.

Let $\lbrace a_{n}\rbrace$ be a sequence of positive terms such that $\sum \limits_{n=1}^{\infty}a_{n}$ diverges.

I am going to show that the series $$\sum \limits_{n=1}^{\infty}\dfrac{a_{n}}{1+a_{n}^{2}}$$ sometimes converges and sometime diverges.

My Attempt: For the divergence of $\sum \limits_{n=1}^{\infty}\dfrac{a_{n}}{1+a_{n}^{2}}$, I defined the sequence $\lbrace a_{n}\rbrace$ as follows.

For each $n\in \mathbb{N}$, $a_{n}=1$. Then $\sum \limits_{n=1}^{\infty}a_{n}=\sum \limits_{n=1}^{\infty}1$ and $\sum \limits_{n=1}^{\infty}\dfrac{a_{n}}{1+a_{n}^{2}}=\sum \limits_{n=1}^{\infty}\dfrac{1}{2}$. Hence both $\sum \limits_{n=1}^{\infty}a_{n}$ and $\sum \limits_{n=1}^{\infty}\dfrac{a_{n}}{1+a_{n}^{2}}$ are divergent. But I am having trouble finding an example for the convergence of $\sum \limits_{n=1}^{\infty}\dfrac{a_{n}}{1+a_{n}^{2}}$.

Can any one please give me a hint or an idea?

• Try something where the denominator $1+a_n^2$ is much larger than the numerator $a_n$. Jan 31, 2015 at 17:49
• What about $a_n = n^2$? (or $a_n = n^b$ with $b \ge 2$) Jan 31, 2015 at 17:49
• $a_n=n$ and $a_n=n^2$ are quite easy to find. Jan 31, 2015 at 18:06

Hint. Let $r>1$. If $a_n>r^n$ then $\sum_{n=1}^{\infty}a_n=\infty$. Note that $$\sum_{n=1}^{\infty} \frac{a_n}{1+a_n^2} = \sum_{n=1}^{\infty} \frac{1}{\frac{1}{a_n}+a_n} \leq \sum_{n=1}^{\infty}\frac{1}{r^n} = \lim_{n\to \infty}\frac{(\frac{1}{r})^{n+1}-\frac{1}{r}}{(\frac{1}{r})-1} = \frac{-\frac{1}{r}}{\frac{1}{r}-1} = \frac{1}{r-1}$$
Let $a_n = 2^n$. Then $\sum_{n = 1}^\infty a_n$ diverges. However, since $a_n/(1 + a_n^2) < 1/2^n$ and $\sum_{n = 1}^\infty 1/2^n$ converges, by the comparison test, $\sum_{n = 1}^\infty a_n/(1 + a_n^2)$ converges.
Consider $a_n = \frac{1}{n^{1+\epsilon}}, \ \epsilon>0$. After you do some algebra the main term in the sequence will become $\frac{1}{n^{1+\epsilon}}$, which converges (e.g. by integral test). Take for example $a_n = \frac{1}{n^2}$.
• Sorry to comment on such an old answer, but the OP wants $\sum a_n$ to diverge. Sep 4, 2018 at 16:30