anonymous
  • anonymous
How do I solve this I know the I am supposed to use Alternating Series Test, but I have not encountered (-2)^n, i am used to (-1)^n http://i.imgur.com/bGGI3yy.png Thanks
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ChillOut
  • ChillOut
You can just rewrite \((-2)^{n}=2^{n}*(-1)^{n}\)
ChillOut
  • ChillOut
Do you know the alternating series test?
anonymous
  • anonymous
It has to satisfy that an+1<= an and lim an = 0

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anonymous
  • anonymous
in order to be convergent
ChillOut
  • ChillOut
Rewrite that series in terms of \(a_{n}=(-1)^{n}*b_{n}\). If \(b_{n}\) decreases and \(lim_{n \rightarrow \infty}b_{n}=0\), \(a_{n}\) converges..
ChillOut
  • ChillOut
Are you having trouble on finding \(b_{n}\)?
anonymous
  • anonymous
is bn just 2^n/7^n
ChillOut
  • ChillOut
What about the n?
ChillOut
  • ChillOut
I mean, there's a n multiplying what you just posted.
anonymous
  • anonymous
Oh right my bad so bn=n2^n / 7^n right?
ChillOut
  • ChillOut
yes. You can just write that as \(n*(\frac{2}{7})^{n}\)
ChillOut
  • ChillOut
Can you finish it?
anonymous
  • anonymous
I am having troubles finding the limit. can i just use l'hopitals rule?
ChillOut
  • ChillOut
Yes! You will need l'hospital's rule to do it. But you need to rewrite it first before applying the rule.
ChillOut
  • ChillOut
Alternatively, you can notice pretty easily that \(7^{n}\) grows WAY faster than \(n*2^{n}\). But as we are doing the "formal" way, just solve the limit.
anonymous
  • anonymous
Ah yes 7^n grows much faster than n*2^n therefore the limit is zero.
ChillOut
  • ChillOut
Yes. You might try doing the limit though.
ChillOut
  • ChillOut
Now to finish the alternating series test you need if the sequence { \(b_{n}\) } decreases. do you know how?
anonymous
  • anonymous
I would use the first derivative test on bn. to See if it eventually decreases
ChillOut
  • ChillOut
You just need to check if \(b_{n}>b_{n+1}\). Your method would work too, but you need to be careful with the intervals!
ChillOut
  • ChillOut
You can now finish the question without problems.
anonymous
  • anonymous
If i took the absolute value of an |an| = would i get n*2^n/7^n
ChillOut
  • ChillOut
Yes.
ChillOut
  • ChillOut
But you just need to check the limit and if the sequence \(b_{n}\) decreases really.
anonymous
  • anonymous
Oh okay. thanks!
ChillOut
  • ChillOut
Oh, right, we need to check for absolute convergence!
ChillOut
  • ChillOut
I missed that. But it seems you got it.
anonymous
  • anonymous
So by taking the absolute value it would no longer be an alternating series?
ChillOut
  • ChillOut
Yes. By taking |an| you check for absolute convergence.

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