Finding radius of convergence on a series

**iva** · August 31st 2011, 10:56 AM

I hope this is the right forum for this, it is a differential equations I'm doing, with series now which I'm reviewing before getting into the calculus of it.

I have to find the radius and interval of convergence for this series, and i'm using a ratio test but I can't get past a certain point. my original equation is:

$\sum_{n=1}^{\infty} \frac {(-1)^{n-1} x^{2n-1}} {(2n-1)!}$

I then applied the ratio test (i.e. n+1 series on top and original series on the bottom) and get this far:

$\lim_{n}^{\infty} \frac{-x}{2n} = \frac{-x}{2}\lim_{n}^{\infty} \frac{1}{n}$

Seeing as its the harmonic series whose sum is infinity i am baffled at how to simplify this so i can continue with my equation to get my radius and interval of conversion.

Any advice really appreciated

Thank you

**Prove It** · August 31st 2011, 11:03 AM

Originally Posted by iva

I hope this is the right forum for this, it is a differential equations I'm doing, with series now which I'm reviewing before getting into the calculus of it.

I have to find the radius and interval of convergence for this series, and i'm using a ratio test but I can't get past a certain point. my original equation is:

$\sum_{n=1}^{\infty} \frac {(-1)^{n-1} x^{2n-1}} {(2n-1)!}$

I then applied the ratio test (i.e. n+1 series on top and original series on the bottom) and get this far:

$\lim_{n}^{\infty} \frac{-x}{2n} = \frac{-x}{2}\lim_{n}^{\infty} \frac{1}{n}$

Seeing as its the harmonic series whose sum is infinity i am baffled at how to simplify this so i can continue with my equation to get my radius and interval of conversion.

Any advice really appreciated

Thank you

You have not applied the ratio test correctly.

By the ratio test, the series will be convergent when $\displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| < 1$ .

**girdav** · August 31st 2011, 11:05 AM

The ratio test is a good idea. Fix $x\neq 0$ and put $a_n:=\frac{(-1)^{n-1}x^{2n-1}}{(2n-1)!}$ . We have to compute $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$ . We get $0$ hence the radius of convergence is infinite.

**iva** · August 31st 2011, 11:15 PM

Thanks guys, was my calculation of the ratio correct though ? because the limit for 1/n as n tends to infinity works out to 0. So I can conclude that there is no radius of convergence as the series is divergent?

**HallsofIvy** · September 1st 2011, 05:26 AM

You did not apply the ratio test correctly for several reasons. First, because the ratio test only applies to series of positive numbers. That is why girdav has the absolute value in $\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|$ . You should not have that "-".

Also, you have done the cancelation wrong. $|a_n|= \frac{x^{2n-1}}{(2n-1)!}$ but $a_{n+1}= \frac{x^{2(n+1)-1}}{(2(n+1)-1)!}= \frac{x^{2n+1}}{(2n+1)!}$ . Now the ratio is
$\left|\frac{a_{n+1}}{a_n}\right|= \frac{x^{2n+1}}{(2n+1)!}\frac{(2n-1)!}{x^{2n-1}}= \frac{x^{2n+1}}{x^{2n-1}}\frac{(2n-1)!}{(2n+1)!}$
$= x^{(2n+1)- (2n-1)}\frac{(2n-1)!}{(2n+1)(2n)(2n-1)!}= \frac{x^2}{(2n+1)(2n)}$

Of course, that still goes to 0 so the "radius of convergence" is infinite- this series converges for all x.

In fact, it is just the MacLaurin series for cos(x).

**iva** · September 1st 2011, 07:11 AM

Gosh, finally got it after almost posting another question and then figuring it out while doing so. Thanks so much!

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