# Alternating Series

• Feb 18th 2009, 12:33 PM
nikie1o2
Alternating Series
Hi everyone,

I have the Series (-4) ^ K / K^2 and have to find if its absolutely convergent, conditionally convergent or divergent. I know the the generic form of alternating series is (-1)^k or (-1)^ k+1 so i wasnt sure how to go about this one. Also the Series cos(pie)k / k..Since its bounded between -1 and 1 can i just make l Uk l =1/k...?

Thanks for your time and help!
• Feb 18th 2009, 01:01 PM
Jester
Quote:

Originally Posted by nikie1o2
Hi everyone,

I have the Series (-4) ^ K / K^2 and have to find if its absolutely convergent, conditionally convergent or divergent. I know the the generic form of alternating series is (-1)^k or (-1)^ k+1 so i wasnt sure how to go about this one. Also the Series cos(pie)k / k..Since its bounded between -1 and 1 can i just make l Uk l =1/k...?

Thanks for your time and help!

For the first series
$\displaystyle \sum_{k=1}^{\infty} \frac{(-1)^k 4^k}{k^2}$

Since
$\displaystyle \lim_{k \to \infty} \frac{4^k}{k^2} \to \infty$

your series will diverge. For the second, considering

$\displaystyle \sum_{k=1}^{\infty} \frac{ \cos k \pi }{k} = \sum_{k=1}^{\infty} \frac{ (-1)^k }{k}$

Since
$\displaystyle \sum_{k=1}^{\infty} \frac{ 1}{k}$ diverges (it's harmonic) you should use the alternating series test.
• Feb 18th 2009, 01:08 PM
Krizalid
• Feb 18th 2009, 01:21 PM
nikie1o2
Quote:

Originally Posted by danny arrigo
For the first series
$\displaystyle \sum_{k=1}^{\infty} \frac{(-1)^k 4^k}{k^2}$

Since
$\displaystyle \lim_{k \to \infty} \frac{4^k}{k^2} \to \infty$

your series will diverge. For the second, considering

$\displaystyle \sum_{k=1}^{\infty} \frac{ \cos k \pi }{k} = \sum_{k=1}^{\infty} \frac{ (-1)^k }{k}$

Since
$\displaystyle \sum_{k=1}^{\infty} \frac{ 1}{k}$ diverges (it's harmonic) you should use the alternating series test.

Thanks you. I also am stuck on $\displaystyle \sum_{k=2}^{\infty} \frac{(-1)^k}{klnk}$ ...(Doh)
• Feb 18th 2009, 01:26 PM
Krizalid
The same, Leibniz test.

Prove first that $\displaystyle a_k=\frac1{k\ln k}$ is a strictly decreasing sequence for all $\displaystyle k>\frac1e.$