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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?

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

3 Answers 3

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Hint. Let $r>1$. If $a_n>r^n$ then $\sum_{n=1}^{\infty}a_n=\infty$. Note that $$ \sum_{n=1}^{\infty} \frac{a_n}{1+a_n^2} = \sum_{n=1}^{\infty} \frac{1}{\frac{1}{a_n}+a_n} \leq \sum_{n=1}^{\infty}\frac{1}{r^n} = \lim_{n\to \infty}\frac{(\frac{1}{r})^{n+1}-\frac{1}{r}}{(\frac{1}{r})-1} = \frac{-\frac{1}{r}}{\frac{1}{r}-1} = \frac{1}{r-1} $$

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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.

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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 $a_n = \frac{1}{n^2}$.

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  • $\begingroup$ Sorry to comment on such an old answer, but the OP wants $\sum a_n$ to diverge. $\endgroup$ Sep 4, 2018 at 16:30

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