# Math Help - covergense questuin mmn11 2b

1. ## covergense questuin mmn11 2b

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

??

2. $\displaystyle \lim_{k\to \infty} \frac{2k+3}{3k-4} = \frac{2}{3}$

So?

3. Originally Posted by transgalactic
$\sum_{k=2}^{infinity}(-1)^{k}\frac{(2k+3)^{k+1}}{(3k-4)^{k-1}}$
is it converges absolutly ,or with condition(dont know the english term),
The term is, in fact, "conditionally convergent".

or is it diverges?

from nth root method of cauchy
the limit of (2k+3)(3k-4) is infinity

??

4. no they have different powers
numerator will be 1
denominator is -1

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

6. 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