User Name Remember Me? Password

 Calculus Calculus Math Forum

 February 26th, 2018, 12:53 PM #1 Newbie   Joined: Aug 2017 From: Norway Posts: 10 Thanks: 0 Series Hey! I'm stuck on this task about series. Determine whether the following series converges or diverges: $\displaystyle \sum\limits_{n=0}^\infty \frac{n}{n^2 \sqrt n +1}$ I've tried to use the ratio test. All help much appreciated. Last edited by deltaX; February 26th, 2018 at 01:00 PM. February 26th, 2018, 01:39 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,837 Thanks: 1480 note $\displaystyle \sum_{n=0}^\infty \dfrac{n}{n^2\sqrt{n}+1} = \sum_{n=1}^\infty \dfrac{n}{n^2\sqrt{n}+1}$ further ... $\dfrac{n}{n^2\sqrt{n}+1} = \dfrac{1}{n\sqrt{n}+\frac{1}{n}} < \dfrac{1}{n^{3/2}}$ note $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^{3/2}}$ is a convergent p-series Thanks from deltaX February 26th, 2018, 02:08 PM   #3
Senior Member

Joined: Oct 2009

Posts: 755
Thanks: 261

Quote:
 Originally Posted by skeeter note $\displaystyle \sum_{n=0}^\infty \dfrac{n}{n^2\sqrt{n}+1} = \sum_{n=1}^\infty \dfrac{n}{n^2\sqrt{n}+1}$ further ... $\dfrac{n}{n^2\sqrt{n}+1} = \dfrac{1}{n\sqrt{n}+\frac{1}{n}} < \dfrac{1}{n^{3/2}}$ note $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^{3/2}}$ is a convergent p-series
Which explains why the ratio test didn't work. The ratio test is just a fancy way of doing a comparison with a geometric series. It can't do comparisons with series as fine as the p-series.

Note that if your series is a fraction of terms in n to some (possibly fractional) power, then a comparison will always work. The limit comparison test is there probably the easiest. Tags series Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post stevewilliams Algebra 1 April 22nd, 2016 01:40 PM king.oslo Complex Analysis 0 December 28th, 2014 06:50 AM g0bearmon Real Analysis 2 May 22nd, 2012 12:10 PM The Chaz Real Analysis 11 February 7th, 2011 04:52 AM g0bearmon Calculus 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      