# Weierstrass M-test

1. Feb 14, 2010

### Kate2010

1. The problem statement, all variables and given/known data

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]

2. Relevant equations

3. 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)krk|$$\leq$$ rk for r<1 and -1<x$$\leq$$0
However, I do not know how to tackle the case when x=-1.

2. Feb 14, 2010

### 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.

3. Feb 14, 2010

### rsa58

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

4. Feb 14, 2010

### 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?

5. Feb 14, 2010

### 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.