SUM (-1)^n * x^n / n(n-1)
Im just learning this topic so bear with me.
You can solve this by using the ratio test.
Absolute value of A(n+1)/A(n)
The Absoulte value of (-1)^n goes away since its just 1.
------- * ---------
cross cancel you get
x limit of n -> infinity (n-1)/(n+1) which ='s 1
abs of x < 1 = -1 < x < 1
then plug -1 and 1 in and test if they converge of not.
The 2nd problem you can solve by doing the root test.
SUM n / 2^n * (x-1)^n
if you root the equation to the n, the power n cancel out
= n/2 * x-1
you can move x-1 outside of the series and have lim n -> infinity n/2 which = infinity only if x is not 1.
If anyone who can double check me would be great. Again, Im not 100% sure, Im still learning this topic.
The third one you can do by using the ratio test again. Try it.