# Math Help - convergence again..... ><

1. ## convergence again..... ><

sum n=1 oo [(-1)^n / n^2/3 + 1 / n^3/2]

so the second part is easy. p series 3/2>1 converges

but i cant get the first part? im too familiar with the alternating series test

((-1)^n) 1 / n^2/3

do you just try to find the limit of that by doing

lim n->oo 1 / n^2/3 = 1/oo = 0?

2. Originally Posted by jeph
sum n=1 oo [(-1)^n / n^2/3 + 1 / n^3/2]

so the second part is easy. p series 3/2>1 converges

but i cant get the first part? im too familiar with the alternating series test

((-1)^n) 1 / n^2/3

do you just try to find the limit of that by doing

lim n->oo 1 / n^2/3 = 1/oo = 0?
You should try to apply the following theorem:
Alternating series test: A series of the form ∑[ (−1)^n] an (with an ≥ 0) is called alternating. Such a series converges if the sequence an is monotone decreasing and converges to 0. The converse is in general not true.

You noticed that bn= 1/n^(2/3) converges to 0.
All you have to do is to show that is also decreasing.
Use a ratio test:bn+1/bn=n^2/3/(n+1)^2/3=(n/n+1)^2/3<1, so decreasing.

3. how do you show (n/n+1)^2/3<1

4. Since n goes from 1 to infinity, is a positive
So, n<n+1,and n^2/3<(n+1)^2/3
Therefore, [n/(n+1)]^2/3<1