Convergent and Divergent Series

[tasveerk]

Homework Statement

For what integer k, k > 1, will both sigma n=1 to infinity ((-1)^(kn))/n and sigma n=1 to infinity (k/4)^n converge?
A) 6 B) 5 C)4 D)3 E)2

Homework Equations

The Attempt at a Solution

I tried to use the ratio test and after some simplifying I got (-1)^k (n/n+1) for the first one and k/4 for the second one. I understand that for the second one any number smaller than 4 will work, but I am unsure of the first one. The answer that is listed is 2.

 

[Char. Limit]

Well, you know that any number smaller than 4 will work. Now I'll ask you, what is the first series for k even? How about k odd?

 

[tasveerk]

Thanks for the quick reply. Is there some type of rule that states if the ratio is negative 1, the series will converge? I thought there were absolute value bars around the ratio meaning that -1 and 1 would be the same thing.

 

[Char. Limit]

Actually, when you have an alternating series like (-1)ⁿ/n, you can use the Alternating Series test, which is far more efficient than a ratio test. As for (-1)²ⁿ/n = 1/n, well that's a series you should recognize.

 

[tasveerk]

I now see that it is a harmonic series, but how did you conclude that (-1)2n/n = 1/n?

 

[Char. Limit]

Well, n is a natural number. And for every natural number n, 2n is even. And -1 to an even power is 1.

 

[tasveerk]

I see what you mean now. I was confusing my results from the ratio test with what you were saying. Just to clarify one last thing, (-1)/n is a convergent series, correct?

 

[Char. Limit]

Yes it is, as you can verify with the Alternating Series test.

 

[tasveerk]

It all makes sense now. Thanks