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anonymous
 one year ago
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
anonymous
 one year ago
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

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ChillOut
 one year ago
Best ResponseYou've already chosen the best response.1You can just rewrite \((2)^{n}=2^{n}*(1)^{n}\)

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.1Do you know the alternating series test?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It has to satisfy that an+1<= an and lim an = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in order to be convergent

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.1Rewrite 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
 one year ago
Best ResponseYou've already chosen the best response.1Are you having trouble on finding \(b_{n}\)?

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.1I mean, there's a n multiplying what you just posted.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh right my bad so bn=n2^n / 7^n right?

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.1yes. You can just write that as \(n*(\frac{2}{7})^{n}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am having troubles finding the limit. can i just use l'hopitals rule?

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.1Yes! You will need l'hospital's rule to do it. But you need to rewrite it first before applying the rule.

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.1Alternatively, 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
 one year ago
Best ResponseYou've already chosen the best response.0Ah yes 7^n grows much faster than n*2^n therefore the limit is zero.

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.1Yes. You might try doing the limit though.

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.1Now to finish the alternating series test you need if the sequence { \(b_{n}\) } decreases. do you know how?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I would use the first derivative test on bn. to See if it eventually decreases

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.1You 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
 one year ago
Best ResponseYou've already chosen the best response.1You can now finish the question without problems.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If i took the absolute value of an an = would i get n*2^n/7^n

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.1But you just need to check the limit and if the sequence \(b_{n}\) decreases really.

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.1Oh, right, we need to check for absolute convergence!

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.1I missed that. But it seems you got it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So by taking the absolute value it would no longer be an alternating series?

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.1Yes. By taking an you check for absolute convergence.
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