Question

determine whether this series converges or diverges sigma 1 to infinity -1^(n-1)2^n/(n^2)

Answer 1

These fractional types usually suggest ratio test

Answer 2

\[\sum_{n=1}^{\infty}\frac{(-1)^{n-1} 2^n}{n^2}\]

as chag said, using the ratio test, you'll get:

\[|\frac{an+1}{an}| = |\frac{(-1)^n2^{n+1}}{(n+1)^2} . \frac{n^2}{(-1)^{n-1}2^n}|\]

Answer 3

so :

Answer 4

\[= \frac{2^n.2^1}{(n+1)^2} . \frac{n^2}{2^n} = \frac{2n^2}{(n+1)^2}\]

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 ^_^

Answer 5

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

Answer 6

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

Answer 7

ty i got it

Answer 8

np ^_^

Answer 9

can u help me with another problem

Answer 10

I'll give it a try ^_^