## anonymous 5 years ago determine whether this series converges or diverges sigma 1 to infinity -1^(n-1)2^n/(n^2)

1. anonymous

These fractional types usually suggest ratio test

2. anonymous

$\sum_{n=1}^{\infty}\frac{(-1)^{n-1} 2^n}{n^2}$ as chag said, using the ratio test, you'll get: $|\frac{an+1}{an}| = |\frac{(-1)^n2^{n+1}}{(n+1)^2} . \frac{n^2}{(-1)^{n-1}2^n}|$

3. anonymous

so :

4. anonymous

$= \frac{2^n.2^1}{(n+1)^2} . \frac{n^2}{2^n} = \frac{2n^2}{(n+1)^2}$ now the second step is to find the limit as n--> infinity: $\lim_{n \rightarrow \infty} \frac{2n^2}{(n+1)^2} = 2$ and the Ratio Test's theorem says the following: if L (limit) > 1 = series diverge if L <1 = series converge if L = 1 , no conclusion Correct me if I'm wrong please ^_^

5. anonymous

so for this case, since 2 > 1, then the following series diverge :)

6. anonymous

if you don't understand what I did, let me know :)

7. anonymous

ty i got it

8. anonymous

np ^_^

9. anonymous

can u help me with another problem

10. anonymous

I'll give it a try ^_^