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