alternating series test?

Question

determine whether the series is absolutely convergent:

The sum from n=1 to n=infinity, (-1) ^n n/(n^2 +1)

in other words, negative one to the power of n times the dividend of n divided by n squared by one.

Answer 1

So you can use the alernating series test because it alternates and Bn>Bn+1 (each term gets smaller). Take the limits of the Bn which is just the abs value of the series and because it goes to zero the sum is convergent. This sum is only conditionall convergent because you can use the integral test on the absolute value of the function, with a u substitution of n^2+1=u and then du=2ndn to prove its a non convergent p series.

Answer 2

Each term smaller than previous term? Let n=5 1/(n^2-6n+10) = 1/(25-30+10) = 1/5 Let n=6 1/(6^2-30+10) = 1/16 1/16 < 1/5 Yes. 2) does the nth term go to 0? lim n-> infinity 1/(n^2-6n+10) =0 The alternating series converges.

Answer 3

It's not absolutely convergant, since you can show that each term would be greater than a similar term in the series n=2 to infinity (2/n), which does not converge.