 My Math Forum Find the radius of convergence, and the interval of convergence, of the series.

 Calculus Calculus Math Forum

 January 23rd, 2016, 10:43 AM #1 Newbie   Joined: Jan 2016 From: sweden Posts: 11 Thanks: 0 Find the radius of convergence, and the interval of convergence, of the series. Power Series: Find the radius of convergence, and the interval of convergence of the series: $\displaystyle \sum_{n=1}^{\infty }\frac{\sqrt{n}(2x+3)^{n}}{n^{2}+1}$ I have tried to solve this problem as following: $\displaystyle \lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_{n}} \right |=\lim_{n\rightarrow \infty}\left | \frac{\sqrt{n+1}(2x+3)^{n+1}.(n^2+1)}{\sqrt{n}(2x+ 3)^n.((n+1)^2+1)} \right |= \left | 2x+3 \right |$ well, I have skipped some steps here because of too many codings. so $\displaystyle -2< x< -1$ And: $\displaystyle x=-1, \sum_{n=1}^{\infty }\frac{\sqrt{n}(1)^n}{n^2+1}=\sum_{n=1}^{\infty }\frac{n^{1/2}(1)^n}{n^2+1}=\sum_{n=1}^{\infty }\frac{n^{1/2}}{n^2(1+\frac{1}{n^2})}=\frac{1}{n^{3/2}}$ Convergence, according to the root test $\displaystyle x=-2, \sum_{n=1}^{\infty }\frac{\sqrt{n}(-1)^n}{n^2+1}=\sum_{n=1}^{\infty }\frac{n^{1/2}(-1)^n}{n^2+1}=\sum_{n=1}^{\infty }\frac{n^{1/2}(-1)^n}{n^2(1+\frac{1}{n^2})}=\frac{(-1)^n}{n^{3/2}}$ Leibzig gives: 1: The term changes sign for each n 2: $\displaystyle a_{n+1}> a_{n}$ 3: $\displaystyle \lim_{n\rightarrow \infty } a_{n}=0$ But my fussy teacher says that everything is ok except my final answer and it should be written in a correct order (mathematically). So could you please look and tell me what is wrong? thank you. Last edited by skipjack; January 23rd, 2016 at 02:26 PM. January 23rd, 2016, 11:05 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 $-2  January 23rd, 2016, 02:34 PM #3 Global Moderator Joined: May 2007 Posts: 6,856 Thanks: 745 At x = -2, it is absolutely convergent (same expression as x = -1), therefore it is convergent. Convergence at both ends gives interval$\displaystyle -2\leq x \leq-1\$. Tags convergence, find, interval, radius, 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 shreddinglicks Calculus 1 November 6th, 2014 10:43 PM hbarnes Calculus 2 July 19th, 2013 08:00 AM dkssudgktpdy Real Analysis 2 April 5th, 2011 11:14 PM naspek Calculus 2 December 10th, 2009 02:34 PM Makino Calculus 1 June 24th, 2009 09:23 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      