i) An example of a partial fraction (after a little bit of work):
1 = n(A+B)+(A-B)
Now, do you set (A+B) equal to 1 or to 0?
I remember something about the power of n coming into play but cannot remember exactly.
From experience I would set (A-B)=1 because that would produce a negative B, but I want to know why we do this.
ii) for the proof question you gave at the end of monday's lecture, i have a small question about "proper" notation.
The proof: If sigma(bn) converges, show that sigma(bn)^n also converges
A possible solution: lim(n->inf)bn=0 since it is convergent. Therefore from some point on bn<1
It follows then that (bn)>(bn)^n for n > 1 and bn < 1. And by the comparison test, since bn converges, it follows that (bn)^n converges as well.
Is this a correct way of going about a proof-style question?
(i). Think of the left-hand side as being 0n+1, and then equate the coefficients in front of the powers of n for the polynomials on both sides: so, the fact that these polynomials (on n) need to be equal, implies 0=A+B (equating the coefficients in front of n1) and 1=A-B (equating the coefficients in front of n0=1).
(ii) There is a gap in your solution: the comparison test works only for positive series, and we have not assumed that bn is > 0. .
Hmm, could you use the root test in following fashion:
which would equal
and from the fact that sigma(bn) converges, the limit is equal to zero. ie,
lim(n->) (bn) = 0
This value is < 1 and by the root test the series is absolutely convergent.
OK, except that instead of lim(n->inf) [(bn)^n]^(1/n), you should write lim(n->inf) |(bn)^n|^(1/n) (and you should continue using the absolute value). Close to the end you need to observe that since lim(n->) (bn) = 0, then lim(n->) |bn| = 0 and your conclusion follows.