# Math Help - Series Convergence

1. ## Series Convergence

So here is the series

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

For what values of x will it converge?

2. ## Re: Series Convergence

According to the ratio test, a series is convergent when \displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \end{align*}.

So get an expression for this limit, set it less than 1, then solve for x.

3. ## Re: Series Convergence

Right I was thinking along the lines of ratio test

I get this

$\lim_{n\rightarrow\infty} \frac{x^2}{2n(2n+1)}$

which becomes
$x^2\lim_{n\rightarrow\infty} \frac{1}{2n(2n+1)}$

clearly, the limit goes to 0 which is less than 1 so it converges. What can i say about x?

4. ## Re: Series Convergence

Since the limit is always 0 no matter the value of x, the series is convergent for all x.

5. ## Re: Series Convergence

ahh I see, I had that feeling too, wanted to be sure

6. ## Re: Series Convergence

For this series,

$\Sigma\limits_{n=1}^\infty \frac{n(x-1)^n}{2^n}$

Clearly by ratio test, I get that

$\lim_{n\rightarrow\infty}{a_{n}}^{1/n} = 1$. So it's inconclusive here. What can I say about which values of x will it converge?

7. ## Re: Series Convergence

Originally Posted by Kanwar245
For this series,

$\Sigma\limits_{n=1}^\infty \frac{n(x-1)^n}{2^n}$

Clearly by ratio test, I get that

$\lim_{n\rightarrow\infty}{a_{n}}^{1/n} = 1$. So it's inconclusive here. What can I say about which values of x will it converge?
If the ratio test is inconclusive you need to use a different test. Maybe the root test?

8. ## Re: Series Convergence

My bad, I meant the root test failed.

EDIT: Ratio test also gives 1

9. ## Re: Series Convergence

Originally Posted by Kanwar245
For this series,

$\Sigma\limits_{n=1}^\infty \frac{n(x-1)^n}{2^n}$

Clearly by ratio test, I get that

$\lim_{n\rightarrow\infty}{a_{n}}^{1/n} = 1$. So it's inconclusive here. What can I say about which values of x will it converge?
Maybe you should try the ratio test then...

\displaystyle \begin{align*} \left| \frac{a_{n+1}}{a_n} \right| &= \left| \frac{\frac{(n+1)(x - 1)^{n+1}}{2^{n+1}}}{\frac{n(x-1)^n}{2^n}} \right| \\ &= \left| \frac{2^n(n+1)(x-1)^{n+1}}{2^{n+1}n(x-1)^n} \right| \\ &= \frac{1}{2}|x - 1| \left| \frac{n + 1}{n} \right| \end{align*}

The "n" stuff goes to 1, but the rest doesn't. You should be able to get an interval for x when you set this entire limit less than 1.

10. ## Re: Series Convergence

Originally Posted by Prove It
According to the ratio test, a series is convergent when \displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \end{align*}.

So get an expression for this limit, set it less than 1, then solve for x.
In order to apply the ratio test, I need to be sure that each term in the sequence is positive right, how would I know that without knowing x?

11. ## Re: Series Convergence

Originally Posted by Kanwar245
In order to apply the ratio test, I need to be sure that each term in the sequence is positive right, how would I know that without knowing x?
Not true at all. For any series, if you can show the series of absolute values is convergent, then so is the original series. What we are doing by taking the absolute value of the ratio in the ratio test is to check the convergence of the absolute value series. When this is convergent, so is the original series.