Question

Classify each series as absolutely convergent, conditionally convergent, or divergent.

Answer 1

How is this divergent? Wouldn't a p-series comparison work ?

Answer 2

http://www.wolframalpha.com/input/?i=sum+%28%28-4^k%29%2F%28k^2%29%29+from+1+to+infinity

As k increases, the value decreases exponentially, since -4^k is over k^2.

(-4^100)/(100^2) = -160693804425899027554196209234116260252220299378279283530+(86/625)

Answer 3

the p-series comparison won't work here because you have a 4^k in the numerator. Just compare 4^k to k^2 to see that 4^k will grow much larger than k^2 will.

Plugging in a few numbers,

4^1=4
1^2=1

4^2=16
2^2=4

4^3=64
3^2=9

So we can see each successive term is getting much larger. Just taking the limit of the absolute value of the quantity as it approaches infinity will give us (after two applications of L'Hopital's rule) that this is definitely divergent.