For what values of p does Naturally this series is the same as I tried to set up

Question

For what values of p does

Naturally this series is the same as

I tried to set up a relation using the ratio test, however I was unable to get anything useful from it.

Explanation / Answer

summation from 0 to infinity (-1)^n/(n^p) well, it converges for any p by the alternating series test if p is at least 0 (the alternating harmonic series would be a common example. it goes to 0. Try it - 1/infinity = 0 (if p > 0) monotone decreasing: abs((-1)^n/(n^p))-abs((-1)^(n+1)/((n+1)^p))>0 which is really: 1/(n^p)-1/(n+1)^p>0 and that's either easily seen or can be done by getting the same denominator: ((n+1)^p-n^p)/(n^p+(n+1)^p)>0 the bottom doesn't matter. ((n+1)^p-n^p if p>0 http://www.wolframalpha.com/input/?i=summation+from+1+to+infinity+%28-1%29^n%2F%28n^.1%29 for a test.

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