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 November 21st, 2011, 08:29 PM #1 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Limit of a series For part (a) I have no problem proving the there converge using the Alternating Series test. For part (b) I'm not too sure the logic to come up with the limit. Adding the first five terms I get: $\sum_{n=1}^5 (-1)^n \dfrac{4}{2n+1}=-\dfrac{4}{3}+\dfrac{4}{5}-\dfrac{4}{7}+\dfrac{4}{8}-\dfrac{4}{11}= -.968$ This would lead me to guess the sum converges to -1, but I put it into wolfram it gives me that the sum is $\pi-4$ if I start the series at n=1, and $\pi$ if I start the series at n=0. How do you know where to innate the series if no starting value is given, and does anyone give me some feedback on the sum of this series? thx
 November 21st, 2011, 09:08 PM #2 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Limit of a series In part (a) it gives the index of summation: n = 0 is where you start.
 November 22nd, 2011, 06:59 AM #3 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Limit of a series [color=#000000]These first series can be easily computed with the help of complex analysis. 1. $\sum_{n=0}^{\infty}\frac{4(-1)^{n}}{2n+1}=-\frac{1}{2}\textrm{Residue}\left(\frac{4\pi\csc(\p i z )}{2z+1};z=-\frac{1}{2}\right)=-\frac{1}{2}\lim_{z\to-\frac{1}{2}}\left(z-\frac{1}{2}\right)\frac{1}{2}\frac{4\pi\csc(\pi z )}{z-\frac{1}{2}}=-\lim_{z\to-\frac{1}{2}}\frac{\pi}{\sin(\pi z )}=\pi$ or simpler$\sum_{n=0}^{\infty}\frac{4(-1)^{n}}{2n+1}=4\underbrace{\sum_{n=0}^{\infty}\fra c{(-1)^{n}}{2n+1}}_{\arctan(1)=\frac{\pi}{4}}=4\frac{\ pi}{4}=\pi$ 2. $\sum_{n=0}^{\infty}\frac{4(-1)^{n}}{2n+1}\left(\frac{1}{2^n}+\frac{1}{3^n}\rig ht)=\sum_{n=0}^{\infty}\frac{4(-1)^{n}}{2n+1}\frac{1}{2^n}+\sum_{n=0}^{\infty}\fra c{4(-1)^{n}}{2n+1}\frac{1}{3^n}=4\sqrt{2}\underbrace{\s um_{n=0}^{\infty}\frac{(-1)^{n}\left(\frac{1}{\sqrt{2}}\right)^{2n+1}}{2n+1 }}_{\arctan\left(\frac{1}{\sqrt{2}}\right)}+4\sqrt {3}\underbrace{\sum_{n=0}^{\infty}\frac{(-1)^{n}\left(\frac{1}{\sqrt{3}}\right)^{2n+1}}{2n+1 }}_{\arctan\left(\frac{1}{\sqrt{3}}\right)}=$ $4\left(\sqrt{2}\arctan\left(\frac{1}{\sqrt{2}\righ t)+\sqrt{3}\arctan\left(\frac{1}{\sqrt{3}}\right)\ right)$ .[/color]
 November 22nd, 2011, 08:09 AM #4 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Limit of a series [color=#000000]3. For $|x|<1$, $\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}=\arctan(x)$ So $\sum_{n=0}^{\infty}\frac{4(-1)^n}{2n+1}\left(\frac{4}{5^n}-\frac{1}{239^n}\right)=4\left(4\sum_{n=0}^{\infty} \frac{(-1)^n}{5^n(2n+1)}-\sum_{n=0}^{\infty}\frac{(-1)^n}{239^n(2n+1)}\right)=4\left(4\sqrt{5}\arctan\ left(\frac{1}{\sqrt{5}\right)-\sqrt{239}\arctan\left(\frac{1}{\sqrt{239}\right)\ right)$ It is obvious that these three series do not converge to the same number, so I think that the second question asks about the convergence of the sequences of the series. When a series converges $\sum_{n=0}^{\infty}a_{n}<\infty\Rightarrow a_{n}\to 0=$, so all three sequences converge to 0.[/color]

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