anonymous
  • anonymous
determine whether this series converges or diverges sigma 1 to infinity -1^(n-1)2^n/(n^2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
These fractional types usually suggest ratio test
anonymous
  • anonymous
\[\sum_{n=1}^{\infty}\frac{(-1)^{n-1} 2^n}{n^2}\] as chag said, using the ratio test, you'll get: \[|\frac{an+1}{an}| = |\frac{(-1)^n2^{n+1}}{(n+1)^2} . \frac{n^2}{(-1)^{n-1}2^n}|\]
anonymous
  • anonymous
so :

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anonymous
  • 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 ^_^
anonymous
  • anonymous
so for this case, since 2 > 1, then the following series diverge :)
anonymous
  • anonymous
if you don't understand what I did, let me know :)
anonymous
  • anonymous
ty i got it
anonymous
  • anonymous
np ^_^
anonymous
  • anonymous
can u help me with another problem
anonymous
  • anonymous
I'll give it a try ^_^

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