# Finding interval/radius of convergence of power series

• May 10th 2009, 05:16 PM
Pinkk
Finding interval/radius of convergence of power series
I'm fairly certain I have the correct answer, but I just want to verify.

$\displaystyle \sum_{k=1}^{\infty}\frac{k^{2}x^{k}}{2\cdot 4\cdot 6\cdot \cdot \cdot (2k)}$

So I used the ratio test:

$\displaystyle \lim_{k\to \infty}|\frac{a_{k+1}}{a_{k}}| = \lim_{k\to \infty}\frac{(k+1)^{2}|x|^{k+1}}{4\cdot 6\cdot 8\cdot \cdot \cdot (2k+2)}\cdot \frac{2\cdot 4\cdot 6\cdot \cdot \cdot (2k)}{k^{2}|x|^{k}}=2|x|\lim_{k\to \infty}\frac{k^{2}+2k+1}{2k^{3}+2k^{2}}<1$

Well, the limit goes to zero, so I end up with:

$\displaystyle 2|x|\cdot 0 < 1$

$\displaystyle 0<1$

So that means the series is convergent for all values of $\displaystyle x$, so the radius of convergence is $\displaystyle \infty$ and the interval of convergence is $\displaystyle (-\infty,\infty)$.
• May 10th 2009, 06:10 PM
Banned for attempted hacking
Absolutely Correct.
• May 10th 2009, 06:11 PM
Prove It
Quote:

Originally Posted by Pinkk
I'm fairly certain I have the correct answer, but I just want to verify.

$\displaystyle \sum_{k=1}^{\infty}\frac{k^{2}x^{k}}{2\cdot 4\cdot 6\cdot \cdot \cdot (2k)}$

So I used the ratio test:

$\displaystyle \lim_{k\to \infty}|\frac{a_{k+1}}{a_{k}}| = \lim_{k\to \infty}\frac{(k+1)^{2}|x|^{k+1}}{4\cdot 6\cdot 8\cdot \cdot \cdot (2k+2)}\cdot \frac{2\cdot 4\cdot 6\cdot \cdot \cdot (2k)}{k^{2}|x|^{k}}=2|x|\lim_{k\to \infty}\frac{k^{2}+2k+1}{2k^{3}+2k^{2}}<1$

Well, the limit goes to zero, so I end up with:

$\displaystyle 2|x|\cdot 0 < 1$

$\displaystyle 0<1$

So that means the series is convergent for all values of $\displaystyle x$, so the radius of convergence is $\displaystyle \infty$ and the interval of convergence is $\displaystyle (-\infty,\infty)$.

Note that

$\displaystyle \frac{k^2 + 2k + 1}{2k^3 + 2k^2} = \frac{(k + 1)^2}{2k(k + 1)}$

$\displaystyle = \frac{k + 1}{2k}$

$\displaystyle = \frac{k}{2k} + \frac{1}{2k}$

$\displaystyle = \frac{1}{2} + \frac{1}{2k}$

As $\displaystyle k \to \infty$, the limit tends to $\displaystyle \frac{1}{2}$. So you're right about the ratio being less than 1 and so the series is convergent. Just looks better :P
• May 10th 2009, 06:19 PM
Soroban
Hello, Pinkk!

You're absolutely correct . . . nice work!

Quote:

$\displaystyle \sum_{k=1}^{\infty}\frac{k^{2}x^{k}}{2\cdot 4\cdot 6\cdot \cdot \cdot (2k)}$
I would do it like this . . .

Note that: .$\displaystyle 2\cdot4\cdot6\hdots (2k) \;=\;2^k(1\cdot2\cdot3\hdots k) \;=\;2^k\,k!$

. . So we have: .$\displaystyle \sum^{\infty}_{k=1}\frac{k^2x^k}{2^k\,k!}$

Ratio: .$\displaystyle \left|\frac{(k+1)x^{k+1}}{2^{k+1}(k+1)!}\cdot\frac {2^k\,k!}{k^2x^l}\right| \;\;=\;\;\left|\frac{2^o}{2^{k+1}}\cdot\frac{x^{k+ 1}}{x^k}\cdot\frac{k!}{(k+1)!} \cdot\frac{k+1)^2}{k^2}\right|$

. . . $\displaystyle = \;\;\frac{1}{2}\cdot|x|\cdot\frac{1}{k+1}\cdot\fra c{(k+1)^2}{k^2} \;\;=\;\;\frac{1}{2}\cdot|x|\cdot\frac{k+1}{k^2}\; \;=\;\;\frac{1}{2}\left(\frac{1}{k}+\frac{1}{k^2}\ right)|x|$

And: .$\displaystyle \lim_{k\to\infty}\left[\frac{1}{2}\left(\frac{1}{k} + \frac{1}{k^2}\right)|x|\right] \;=\;0 \quad\hdots\;\text{ etc.}$

• May 10th 2009, 06:23 PM
Pinkk
@Prove It

Isn't it actually:

$\displaystyle \lim_{k\to \infty}\frac{k+1}{k^{2}}=0$

If the limit somehow equaled $\displaystyle \frac{1}{2}$, then the radius of convergence would not be $\displaystyle \infty$.
• May 10th 2009, 06:25 PM
Prove It
Quote:

Originally Posted by Pinkk
@Prove It

Isn't it actually:

$\displaystyle \lim_{k\to \infty}\frac{k+1}{k^{2}}=0$

If the limit somehow equaled $\displaystyle \frac{1}{2}$, then the radius of convergence would not be $\displaystyle \infty$.

Oops, you are right. It should have been $\displaystyle \frac{k + 1}{2k^2}$.

So you're right, that limit is not $\displaystyle \frac{1}{2}$.

Still tends to a value less than 1 though.

$\displaystyle \frac{k + 1}{2k^2} = \frac{k}{2k^2} + \frac{1}{2k^2}$

$\displaystyle = \frac{1}{2k} + \frac{1}{2k^2}$.

So as $\displaystyle k \to \infty$ the limit tends to $\displaystyle 0$.

Still converges though.