∞

SUM (-1)^n * x^n / n(n-1)

n=2

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.

so its:

x^n+1 -----n(n-1)

------- * ---------

n+1(n) -----x^n

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

n=1

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.