Question

k=1 sigma k tends to infinity (-3/2)^(k+1) the seires is converge or diverge how can i indentify this series is converge or diverge.

Answer 1

ratio test?

Answer 2

no ..what is ratio test:(

Answer 3

if the series is \(\{a_n\}\) then the series converges absolutely (and hence also converges) if\[\lim_{n\to\infty}\left|{\{a_{n+1}\}\over\{a_n\}}\right|<1\]

Answer 4

this seris is not geometric series..?

Answer 5

I suppose, I just thought of one way to test for convergence
but we can do it the geometric way...

Answer 6

we can reewrite this\[\sum_{k=1}^\infty(-\frac32)^{k+1}=\sum_{k=1}^\infty(-\frac32)^2(-\frac32)^{k-1}\]in general, geometric series of the form\[\sum_{k=1}^\infty ar^{k-1}\]converge when \(|r|<1\) we get our answer

Answer 7

we could easily find what it actually converges to as well, but that seems not to be required

Answer 8

how can i find this seris sums

Answer 9

heloow turing test r u ther..?

Answer 10

ny one can see my question..?