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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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
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You can just rewrite \((-2)^{n}=2^{n}*(-1)^{n}\)
Do you know the alternating series test?
It has to satisfy that an+1<= an and lim an = 0

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in order to be convergent
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..
Are you having trouble on finding \(b_{n}\)?
is bn just 2^n/7^n
What about the n?
I mean, there's a n multiplying what you just posted.
Oh right my bad so bn=n2^n / 7^n right?
yes. You can just write that as \(n*(\frac{2}{7})^{n}\)
Can you finish it?
I am having troubles finding the limit. can i just use l'hopitals rule?
Yes! You will need l'hospital's rule to do it. But you need to rewrite it first before applying the rule.
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.
Ah yes 7^n grows much faster than n*2^n therefore the limit is zero.
Yes. You might try doing the limit though.
Now to finish the alternating series test you need if the sequence { \(b_{n}\) } decreases. do you know how?
I would use the first derivative test on bn. to See if it eventually decreases
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!
You can now finish the question without problems.
If i took the absolute value of an |an| = would i get n*2^n/7^n
Yes.
But you just need to check the limit and if the sequence \(b_{n}\) decreases really.
Oh okay. thanks!
Oh, right, we need to check for absolute convergence!
I missed that. But it seems you got it.
So by taking the absolute value it would no longer be an alternating series?
Yes. By taking |an| you check for absolute convergence.

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