Patrick Corn
Math 185, 7/13/05
Solution Set 5
Problem 1: If |z|<min(R1, R2), then P∞
n=0 anznand P∞
n=0 bnznboth converge, so then
P∞
n=0(an+bn)znconverges as well. So the radius of convergence of this series is at least min(R1, R2).
(You can surely work out what this means when one or both of the radii is ∞.)
And for the second part, if Ris the radius of convergence of P∞
n=0 anbnzn, we have (by property
(ii) on p. 52):
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R= lim sup
n→∞ anbn≤1
R1
1
R2
so R≥R1R2(provided this product is well-defined).
Problem 2: It is easier to show that P∞
n=1 nanznhas the same radius of convergence; it should
be clear that dividing a series (with no constant term) by zdoes not affect its radius of convergence,
so this is good enough. Suppose the radius of convergence of P∞
n=0 anznis R.
This follows directly from the following computation:
lim sup
n→∞ |nan|1/n = lim
n→∞ n1/n ·lim sup
n→∞ |an|1/n = 1 ·1
R=1
R
by property (ii) on p. 52. (I’m omitting the proof that n1/n →1 here, but you should probably
have put it on your homework, to remind yourself and the reader that you know why it’s true–good
old L’Hopital.)
Problem 3: (a) 1, by the ratio test.
(b) If we replaced z3nby zn, the answer would be 27, by the ratio test:
lim
n→∞
cn
cn+1
= lim
n→∞
(n!)3(3n+ 3)!
(3n)!((n+ 1)!)3= lim
n→∞
(3n+ 3)(3n+ 2)(3n+ 1)
(n+ 1)3= 27.
Now, if P∞
n=0 anznhas radius of convergence R, then P∞
n=0 anz3nhas radius of convergence R1/3
(compare exercise V.13.3), which you can check by noting that it converges for |z3|< R and diverges
for |z3|> R. So the final answer is 3.
(c) It is not hard to see that the radius is the reciprocal of
lim sup
n→∞ 1
n1/n!
= lim sup
n→∞ 1
n1/n!1/(n−1)!
,
but the thing in parentheses goes to 1 as n→ ∞, and the exponent goes to 0, so this sequence has
a limit of 1, which must equal its lim sup. Then the answer is 1/1 = 1.
(d) It should be clear that the sequence (n!)1/n!is just a subsequence of n1/n, which converges
to 1, so it must also converge to 1. So R= 1.
(e) Well, we want the reciprocal of
lim sup
n→∞ (nn)1/n2= li m sup
n→∞ n1/n = 1,
so R= 1.
Problem 4: This is a tricky one. Let z=eiθ. Assume that θ∈(0,2π). Then the real part
of the series is P∞
n=0
cos(nθ)
n, and we’d like to show that this converges if θis not a multiple of 2π.
The proof when we replace cos by sin will be similar.
Anyway, we will use a criterion for convergence due to Dirichlet: Suppose the partial sums of
a series Pbnare all bounded, and suppose that cnis a decreasing sequence of positive numbers
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