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Let $\langle a_n\rangle$ be a sequence of positive numbers.

Consider $\sum_{n=1}^\infty{Y_n}=\sum_{n=1}^\infty{\left(\frac{a_n}{n}+\frac{n}{a_n^2}+\frac{a_n}{n^3}\right)}$. Show this diverges using the comparison test.

I tried to find a $\langle X_n\rangle$ where $\sum_{n=1}^\infty{X_n}$ diverges and $X_n$ is greater than than the given $Y_n$ in the series for each $n$ greater than or equal to some $n_0$ . But couldn't find one...

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By the AM-GM inequality,

$$\frac{a_n}{n} + \frac{n}{a_n^2} + \frac{a_n}{n^3} \ge 3\sqrt[3]{\frac{a_n}{n}\cdot \frac{n}{a_n^2}\cdot \frac{a_n}{n^3}} = \frac{3}{n}.$$

Since $\sum\limits_{n = 1}^\infty \frac{3}{n}$ diverges, so does the series $\sum\limits_{n = 1}^\infty (\frac{a_n}{n} + \frac{n}{a_n^2} + \frac{a_n}{n^3})$ by comparison.

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  • $\begingroup$ Just what I was thinking. $\endgroup$ – marty cohen May 22 '15 at 20:44
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I'm not sure if this qualifies as a comparison test, but the first two terms within your parantheses, i.e. $a_n/n + n/a_n^2$ is enough to establish divergence. Note that since all your terms are positive, you can split up the series into two series. if $\sum_n n/a_n^2$ converges then $a_n$ is bounded below by a positive constant (otherwise $n / a_n^2$ would occasionally be arbitrarily large, infinitely many times). However then $\sum_n a_n/n $ diverges by comparison to the harmonic series.

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