# Power Series

• June 17th 2013, 05:59 PM
jjtjp
Power Series
Here's the problem:

$\sum_{n=1}^{\infty}\frac{(-1)^n4^nx^n}{\sqrt{n}+10}$

so I use ratio test (obviously):

$\lim_{n\rightarrow \infty}\left | \frac{(-1)^{n+1}4^{n+1}x^{n+1}}{\sqrt{n+1}+10}\times \frac{\sqrt{n}+10}{(-1)^n4^nx^n}\right |$

after some cancellations I have:

$\lim_{n\rightarrow \infty}\left | \frac{-4x(\sqrt{n}+10)}{\sqrt{n+1}+10}\right |$

It seems that I could say that $\sqrt{n}$ cancels, but then again, n + 1 will increase faster than n.

I can't figure out where to go from here though. Help? Thanks!
• June 17th 2013, 06:33 PM
Soroban
Re: Power Series
Hello, jjtjp!

Quote:

Here's the problem:

$\sum_{n=1}^{\infty}\frac{(-1)^n4^nx^n}{\sqrt{n}+10}$

So I use ratio test (obviously):

$\lim_{n\rightarrow \infty}\left | \frac{4^{n+1}x^{n+1}}{\sqrt{n+1}+10}\times \frac{\sqrt{n}+10}{4^nx^n}\right |$ . We are taking absolute values; drop the (-1)^n.

After some cancellations I have:

$\lim_{n\rightarrow \infty}\left | \frac{4x(\sqrt{n}+10)}{\sqrt{n+1}+10}\right |$

It seems that I could say that $\sqrt{n}$ cancels,
but then again, n + 1 will increase faster than n. . This is not true.

I can't figure out where to go from here though. .Thanks!

We have: . $4x\cdot\frac{\sqrt{n}+10}{\sqrt{n+1} + 10}$

Divide numerator and denominator by $\sqrt{n}:$

. . $4x\cdot\frac{\frac{\sqrt{n}}{\sqrt{n}} + \frac{10}{\sqrt{n}}}{\frac{\sqrt{n+1}}{\sqrt{n}} + \frac{10}{\sqrt{n}}} \;=\;4x\cdot\frac{\sqrt{\frac{n}{n}} + \frac{10}\sqrt{n}}{\sqrt{\frac{n+1}{n}} + \frac{10}{\sqrt{n}}} \;=\;4x\cdot\frac{1 + \frac{10}{\sqrt{n}}}{\sqrt{1\!+\!\frac{1}{n}} + \frac{10}{\sqrt{n}}}$

Therefore: . $\lim_{n\to\infty} \left[4x\cdot\frac{1 + \frac{10}{\sqrt{n}}}{\sqrt{1\!+\!\frac{1}{n}} + \frac{10}{\sqrt{n}}}\right] \;=\;4x\cdot\frac{1+0}{\sqrt{1\!+\!0} + 0} \;=\;4x$
• June 17th 2013, 06:44 PM
jjtjp
Re: Power Series
Thanks so much! I wish I had paid more attention in high school, seems like the trivial stuff always trips me up.