# Series Question

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• Oct 27th 2008, 09:14 PM
RedBarchetta
Series Question
$\displaystyle \sum\limits_{n = 1}^\infty {\frac{{5\ln n}} {{\sqrt {n^3 } }}}$

Is there any easy way to find the convergence/divergence of this series?

So far I tried using the integral test and came up with a convergence at 20.
This involves integration by substitution. Is there any easier method?
• Oct 28th 2008, 12:31 AM
flyingsquirrel
Hello,
Quote:

Originally Posted by RedBarchetta
Is there any easier method?

Maybe : $\displaystyle \sqrt{n^3}=n^\frac32=n^\frac54n^\frac14$ hence

$\displaystyle 5\sum_{n = 1}^\infty\frac{\ln n}{\sqrt {n^3 }}=5\sum_{n = 1}^\infty\frac{1}{n^\frac54}\times\frac{\ln n}{n^\frac14}$.

As $\displaystyle \frac{\ln n}{n^\frac14}\xrightarrow[n\to\infty]{}0$ one has, for $\displaystyle n$ sufficiently large, $\displaystyle \left|\frac{\ln n}{n^\frac14}\right|\leq 1$ hence $\displaystyle \left|\frac{1}{n^\frac54}\times\frac{\ln n}{n^\frac14}\right|\leq\left|\frac{1}{n^\frac54}\ right|\times 1 = \left|\frac{1}{n^\frac54}\right|$ that is to say $\displaystyle \left|\frac{\ln n}{\sqrt {n^3 }}\right|\leq\left|\frac{1}{n^\frac54}\right|$. As $\displaystyle \sum\frac{1}{n^\frac54}$ is an absolutely convergent series, the series $\displaystyle 5\sum_{n = 1}^\infty\frac{\ln n}{\sqrt {n^3 }}$ converges too. (comparison test)
• Oct 28th 2008, 12:44 AM
RedBarchetta
Thanks.