# covergense questuin mmn11 2b

• Mar 2nd 2011, 01:45 AM
transgalactic
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

??
• Mar 2nd 2011, 03:27 AM
Ted
$\displaystyle \lim_{k\to \infty} \frac{2k+3}{3k-4} = \frac{2}{3}$

So?
• Mar 2nd 2011, 03:55 AM
HallsofIvy
Quote:

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".

Quote:

or is it diverges?

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

??
• Mar 2nd 2011, 04:27 AM
transgalactic
no they have different powers
numerator will be 1
denominator is -1
• Mar 3rd 2011, 04:50 AM
Ted
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 $\displaystyle \sum_{k=2}^{\infty} \frac{(2k+3)^{k+1}}{(3k-4)^{k-1}}$ ?
• Mar 3rd 2011, 06:25 AM
transgalactic
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