# Determine the Interval of Convergence

• Nov 12th 2008, 10:54 PM
kl.twilleger
Determine the Interval of Convergence
Determine the interval of convergence

$\sum \frac{-1^k(x+\frac{\pi}{2})^{2k+1}}{2k+1}$

I used the ratio test and got it down to

$\frac{1}{2}\lim= -(x+\frac{\pi}{2})^3$

but I think I messed up somewhere or maybe I shouldn't use the ratio test and maybe try something different.
• Nov 13th 2008, 09:12 AM
o_O
Is this your series: $\sum_{k=0}^{\infty} \frac{(-1)^k (x+\frac{\pi}{2})^{2k+1}}{2k+1}$

Using the ratio test:
\begin{aligned} \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| & = \lim_{k \to \infty} \left|\frac{ (-1)^{k+1} (x+\frac{\pi}{2})^{2(k+1)+1} }{2(k+1)+1} \ \cdot \ \frac{2k+1}{(-1)^k (x + \frac{\pi}{2})^{2k+1} } \right| \\ & = \lim_{k \to \infty} \left| \frac{ (x+\frac{\pi}{2})^{2k+3} }{2k+3} \ \cdot \ \frac{2k+1}{(x+ \frac{\pi}{2})^{2k+1} }\right| \end{aligned}
\begin{aligned} {\color{white}\lim_{k \to \infty} \left| \frac{a_{n+1}}{a_n} \right|} & = \lim_{k \to \infty} \left| (x+\frac{\pi}{2})^2 \cdot \frac{2k+1}{2k+3} \right| \\ & = \left(x+ \frac{\pi}{2}\right)^2 \cdot \underbrace{\lim_{k \to \infty} \left|\frac{2k+1}{2k+3}\right|}_{=1} \end{aligned}

So by the ratio test, your series absolutely converges and thus converges if $\lim_{k \to \infty} \frac{a_{k+1}}{a_k} = \left(x + \frac{\pi}{2}\right)^2 {\color{red}<} \ 1$

Simply solve.
• Nov 13th 2008, 12:50 PM
Mathstud28
Quote:

Originally Posted by kl.twilleger
Determine the interval of convergence

$\sum \frac{-1^k(x+\frac{\pi}{2})^{2k+1}}{2k+1}{\color{red}=\su m{a_k}}$

I used the ratio test and got it down to

$\frac{1}{2}\lim= -(x+\frac{\pi}{2})^3$

but I think I messed up somewhere or maybe I shouldn't use the ratio test and maybe try something different.

You don't have to use the Ratio test. Just note that $a_k\sim\left[\left(x+\frac{\pi}{2}\right)^2\right]^k$

Which is a geometric series and converges iff the value raised to the n is less then one.

P.S. Dont forget to check the endpoints of your IOC

Also note that for every x that makes this seris converge it is euqal to $\arctan\left(x+\frac{\pi}{2}\right)$
• Nov 13th 2008, 12:58 PM
Jameson
Quote:

Originally Posted by Mathstud28
You don't have to use the Ratio test. Just note that $a_k\sim\left[\left(x+\frac{\pi}{2}\right)^2\right]$

What test is "needed" I think really depends on the level of rigorousness expected for the problem. Your observation about a_k is a good intuitive insight but without justification it's useless in the context of a proof. I'm sure you can easily show why what you said is true but like I said it all depends on the rigorousness required of a solution.
• Nov 13th 2008, 01:01 PM
Mathstud28
Quote:

Originally Posted by Jameson
What test is "needed" I think really depends on the level of rigorousness expected for the problem. Your observation about a_k is a good intuitive insight but without justification it's useless in the context of a proof. I'm sure you can easily show why what you said is true but like I said it all depends on the rigorousness required of a solution.

Yes, I understand what you are saying. A lot of the things I say similar to this are not meant to be taken as solutions, or even as what I would put down. They are more a way to rationalize why the answer is what it is. It also enables the student to form a more intuitive insight into mathematics as a whole. So yes you are correct, this should never be given as a solution especially in a formal situation like a quiz or test but I beleive that making these "intuitive insights" is the key to advanced mathematical success.