For what values of p is each alternating series convergent?

Question

E { [(-1)^(n-1)] / (n^p) }

E { [(-1)^(n-1)] ( [ln(n)]^p) / n }



PLEASE WALK ME THROUGH THIS!!!! DON'T JUST GIVE ME THE ANSWER, I NEED TO SEE WORK OR I WILL NOT PICK YOUR ANSWER AS BEST

Answer 1

1) First the limit of absolute value must be 0
lim (n-1)/n^p = lim 1/n^(p-1) -1/n^p so p
must be >1
Let´se if it is monotone decreasing
For this we can take f(x)= (x-1)/x^p find the derivative and se if it is negative for x=>+infinity
f´(x)= 1/x^2p( x^p-(px^(p-1)(x-1)

the sign depends of x^p(1-p)+p(x^(p-1)=
x^(p-1)( x(1-p)+p) which ,as p>1 ,is negative
So by the criterion of Leibnitz for p>1 the series is convergent(conditionally) and for p>2 absolutely convergent
2)I a_nI must have limit 0 so (1-1/n)*[ln(n)]^p
must=>to 0.As the first factor has limit 1 the 2nd must=>to0 so p must be<0 let´s call -p=q so q>0
We must se if it is monotone decreasing

I a_nI =(1-1/n)/[ln(n)]^q
Take f(x)= (1-1/x)/[ln(x)]^q and study the sign if the derivative.
That I leave it to you

Answer 2

Alternating P Series

Answer 3

By the alternating series test, you only need that the terms go to 0 as n-->infinity. For the first series, this happens if p>0. For the second, this always happens.