Question

Find the exact value sum of each of the following sums. If the series diverges, state so and justify your answer.

Answer 1

\[(a) \sum_{k=0}^{\infty}(-1)^{k}\frac{ 2^{k} }{ 9^{k}}\]

Answer 2

Just to confirm, the fraction is (2^k)/(9^k) ?

Answer 3

Yes.

Answer 4

So this is an alternating series because of the (-1)^k term. So the condition of convergence is that the limit of a sub k as k -> ∞ is 0, and also that a sub k + 1 is less than or equal to a sub k for all k. So starting with the first condition. I can rewrite (2^k)/(9^k) as simply (2/9)^k. As k approaches infinity, the limit does approach 0, so the first condition is satisfied. The second condition is also true because as k increases, the value decreases, meaning a sub k + 1 will definitely be less than a sub k. As far as the exact sum, the summation can be rewritten to mimic a geometric series of the form ar^k. So rewriting in that form gives me (-2/9)^k, where a = 1, r = -2/9. As long as the abs value of r is between 0 and 1, I can say the exact sum of a geometric series is equal to a/(1-r). Therefore I would have 1/(1-(-2/9)) which = 9/11

Answer 5

What about: \[\sum_{k=0}^{\infty}\frac{ (\ln (2))^{2k} }{ (2k)! }\]

Answer 6

Am I looking for an exact sum here as well, or just a statement about the convergence or divergence?

Answer 7

Also looking for the exact sum.

Answer 8

Starting off with convergence or divergence, I would like to try to test using the ratio test first, as it usually is a good test to try. So I need to check the limit as k -> ∞ of abs|(a sub k +1)/(a sub k)|. So a sub k + 1 would be (ln(2))^[2(k+1)]/2(k+1)! This is then divided by the original a sub k. Simplifying a sub k + 1 and then dividing by a sub k, I get abs| [(ln(2))^2k*(ln(2))^2*(2k!)] / [2(k+1)(k!)*(ln(2))^2k] | After cancelling, I am left with abs|(ln(2))^2/(k+1)|. As limit of k -> ∞, I get a limit of 0. By the conditions of the ratio test, if abs|(a sub k + 1)/(a sub k)| has a limit less than 1, I have convergence. Admittedly, I am a bit rusty on some of these, so I will send this for now as to not keep you waiting. I'm working out the exact sum part, but this is the proof of convergence for now :3

Answer 9

Thank you!