Asked by Ashley
Show whether the series ((-1)^n)*((2^n)/(n^4))) is absolutely convergent.
Taking an=((-1)^n)*((2^n)/(n^4))) , here |an|=((2^n)/(n^4)).
I have the gut feeling that this is absolutely convergent. But don't see a way to do that. Also, we can only use comparison test, monotone convegence theorem, convergent==>boundedness, convergent sequence has only one limit.
To show |an| is absolutely convergent(which will also prove an is convergent), I need to find another bn, which converges, such that an<=bn. I know bn=n!/n^4 will do, but need to prove that converges first. How can I prove this?
Maybe we can use comparison test seperately more than once right?
Any thoughts on solving this?
Thanks!
Taking an=((-1)^n)*((2^n)/(n^4))) , here |an|=((2^n)/(n^4)).
I have the gut feeling that this is absolutely convergent. But don't see a way to do that. Also, we can only use comparison test, monotone convegence theorem, convergent==>boundedness, convergent sequence has only one limit.
To show |an| is absolutely convergent(which will also prove an is convergent), I need to find another bn, which converges, such that an<=bn. I know bn=n!/n^4 will do, but need to prove that converges first. How can I prove this?
Maybe we can use comparison test seperately more than once right?
Any thoughts on solving this?
Thanks!
Answers
Answered by
oobleck
exponentials grow faster than powers.
For sufficiently large n, 2^n/n^4 >1
and things explode.
and n doesn't even have to be very big.
For sufficiently large n, 2^n/n^4 >1
and things explode.
and n doesn't even have to be very big.
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