$\displaystyle \sum_{k=2}^{infinity}(-1)^{k}\frac{(2k+3)^{k+1}}{(3k-4)^{k-1}}$

is it converges absolutly ,or with condition(dont know the english term),

or is it diverges?

from nth root method of cauchy

the limit of (2k+3)(3k-4) is infinity

??

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- Mar 2nd 2011, 12:45 AMtransgalacticcovergense questuin mmn11 2b
$\displaystyle \sum_{k=2}^{infinity}(-1)^{k}\frac{(2k+3)^{k+1}}{(3k-4)^{k-1}}$

is it converges absolutly ,or with condition(dont know the english term),

or is it diverges?

from nth root method of cauchy

the limit of (2k+3)(3k-4) is infinity

?? - Mar 2nd 2011, 02:27 AMTed
$\displaystyle \displaystyle \lim_{k\to \infty} \frac{2k+3}{3k-4} = \frac{2}{3}$

So? - Mar 2nd 2011, 02:55 AMHallsofIvy
- Mar 2nd 2011, 03:27 AMtransgalactic
no they have different powers

numerator will be 1

denominator is -1 - Mar 3rd 2011, 03:50 AMTed
I did not say anything about your series.

I've just answered the question which you edited it.

In order to test the series for absolutely/condintionally convergent, you should check the absolute series.

Did try anything for $\displaystyle \displaystyle \sum_{k=2}^{\infty} \frac{(2k+3)^{k+1}}{(3k-4)^{k-1}}$ ? - Mar 3rd 2011, 05:25 AMtransgalactic
first we need to test the absolute value series

the n'th root doesnt work because i get a limit of infinity

what you did what takng the expression witout the powers on numerator and denominator