Does the Alternating Series Converge?

[Heat]

I need help identifying if it converges or diverges or conditionally converges.

$$\Sigma$$$$(-1)^{k}$$$$\frac{(k+4)}{(k^{2}+k)}$$


First I want to test for absolute convergence, and comparing this limit to 1/k I get that it diverges. Since it diverges, I need to test it now using the Alternating Series test, in which the limit is 0 and it converges. So my question is, does it conditionally converge?

 

[HallsofIvy]

So your question is just whether the sequence (k+4)/(k²+ k) converges to 0 as k goes to infinity? Divide both numerator and denominator by k². Now what happens as k goes to infinity?

 

[Heat]

yes from 0 to infinity.

it would be (1/k + 4/k)(1+1/k)...4/1 = 4...so?

 

[snipez90]

Ok, you've established that it does not absolutely converge so you must test for conditional convergence, but your test is not comprehensive. Designate a_sub_n as (k+4)/(k^2 + k) and taking HallsofIvy's suggestion, we have lim(n -> +inf) of a_sub_n = lim(n -> +inf) of (1/k + 4/k^2)/(1 + 1/k), which tends to 0 because all the terms with k in the denominator go to 0. Now you have to show that a_sub_(n+1) < a_sub_n, and if this is true, then you can say that the series conditionally converges.

 

[Pere Callahan]

yes from 0 to infinity.

it would be (1/k + 4/k)(1+1/k)...4/1 = 4...so?

This is not correct. Note, that the 4 in the numerator comes as 4/k ...in fact it should be 1/k^2.