Question

determine convergence or divergence. for converging series, state absolute or conditional (explain why):
sum(k=2 -> infinity) [(-1)^k/(k^2-1)]

Answer 1

you can actually find the sum

Answer 2

though a limit comparison test will tell you quickly it converges

Answer 3

Use alternating series test to say:
\[a_{k+1}<a_k \]
For sufficiently large n.
\[a_k \ge 0\]
For all n.
\[\lim_{k \rightarrow \infty} a_k \rightarrow 0\]
Therefore it converges by the alternating series test. So it is conditionally convergent. Now to find it it is absolutely convergent consider the summation:
\[\sum_{k=2}^{\infty} \frac{1}{k^2-1}\]
By limit comparison with \[\sum_{k=2}^{\infty} \frac{1}{k^2}\]
This series converges so we can say that the original series converges based on the limit comparison test.

Answer 4

Sufficiently large k.
For all k.

Not "n".

Answer 5

thanks.. can you help me out with finding if it converges absolutely or conditionaly

Answer 6

Well, conditionally means it converges ONLY with the (-1)^k, and absolutely means it converges regardless. So since it converges in both cases it converges absolutely.

Answer 7

@ Zarkon

Answer 8

@Zarkon

Answer 9

yes?

Answer 10

can you help me out with finding if it converges absolutely or conditionaly

Answer 11

it is absolutely convergent by the limit comparison test

Answer 12

is it because the absolute value of zero is zero?

Answer 13

no

Answer 14

oops, is it because the absolute value of (-1)^k/(k^2-1) is also zero

Answer 15

\[\lim_{k\to\infty}\frac{k^2}{k^2-1}=1\]

the limit is positive finite so

\[\sum_{k=2}^{\infty}\frac{1}{k^2-1}\]

converges since \[\sum_{k=2}^{\infty}\frac{1}{k^2}\] converges

Answer 16

malevolence19 provided you with what you needed...though there is no need for the alternating series test (in this case)

Answer 17

thanks..