Finding interval/radius of convergence of power series

**Pinkk** · May 10th 2009, 05:16 PM

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

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

So I used the ratio test:

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

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

$2|x|\cdot 0 < 1$

$0<1$

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

**Banned for attempted hacking** · May 10th 2009, 06:10 PM

Absolutely Correct.

**Prove It** · May 10th 2009, 06:11 PM

Originally Posted by Pinkk

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

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

So I used the ratio test:

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

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

$2|x|\cdot 0 < 1$

$0<1$

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

Just a note about your ratio test.

Note that

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

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

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

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

As $k \to \infty$ , the limit tends to $\frac{1}{2}$ . So you're right about the ratio being less than 1 and so the series is convergent. Just looks better :P

**Soroban** · May 10th 2009, 06:19 PM

Hello, Pinkk!

You're absolutely correct . . . nice work!

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

I would do it like this . . .

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

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

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

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

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

**Pinkk** · May 10th 2009, 06:23 PM

@Prove It

Isn't it actually:

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

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

**Prove It** · May 10th 2009, 06:25 PM

Originally Posted by Pinkk

@Prove It

Isn't it actually:

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

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

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

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

Still tends to a value less than 1 though.

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

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

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

Still converges though.

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