Weierstrass M-test Homework: 0<p<1

[Kate2010]

Homework Statement

0<p<1
Suppose $$\sum$$$$^{infinity}_{k=0}$$ p(p-1)...(p-k+1)(-1)^k/k(k-1)...1 is convergent.
Show that $$\sum$$$$^{infinity}_{k=0}$$ p(p-1)...(p-k+1)(x)^k/k(k-1)...1 is uniformly convergent on [-1,0]

Homework Equations

The Attempt at a Solution

I have shown that p(p-1)...(p-k+1)(-1)^k/k(k-1)...1 < 0 for k=1,2,3,...
$$\sum$$$$^{infinity}_{k=0}$$ p(p-1)...(p-k+1)(-1)^k/k(k-1)...1 = L (< 0) as it converges to a limit.
|(-1)^kr^k|$$\leq$$ r^k for r<1 and -1<x$$\leq$$0
However, I do not know how to tackle the case when x=-1.

 

[rsa58]

remember the wierstrass m-test holds for the absolute value of the functions in the sequence. furthermore the first series is not strictly negative it is strictly positive this has to do with the parity between the p-k terms and (-1)^k. therefore the m-test really is applicable. i.e. the absolute value of the terms in the second series really are less than the corresponding terms in the first. hence uniform convergence.

 

[rsa58]

note as you mentioned for the case x=-1 the terms are equal and this is acceptable condition for the m-test

 

[Kate2010]

I'm pretty sure the 1st series is strictly negative as it was a show that question. Could I just consider the negative of that series?

 

[rsa58]

you can. sorry i misread the sum. yeah if you multiply by negative -1 the series converges by m-test. then since the negative of the series converges then a constant multiple of the series (by -1) also converges to the negative of the limit.