The University of Waikato Department of Mathematics Elements of Analysis Worshop/Tutorial for 1st October 2008: math252-08B. Attempt questions 1-5.
Name:|||||||||||
The limit supremum:
1. For each n 2 N let an = ( 1)n+ 1=n. Show that 2 an 2 for all n, showing that the sequence is bounded. Evaluate
bn= supfan; an+1; g
and show it is decreasing with n. Find the limit limn!1bn = lim supn!1an. Finally make a sketch of the rst 4 terms of an and bn on the same axes.
2. Do the same for cn= 1 + 1=n and verify that, in this case, lim sup
n!1 cn= lim
n!1cn:
3. Show that (n!)1=n ! 1 as n ! 1 by rst establishing the bound n! (n=2)n=2.
The radius of convergence:
4. Use the limit formula
Rf = lim
n!1 an an+1 to nd the radius of convergence of
f(x) = 1 + mx + m(m 1)
2! x2+ +m(m 1)( )(m n + 1)
n! xn+
for m 2 R.
5. Show why the limit formula does not work for cos(x) = 1 x2
2! + + ( 1)nx2n 2n! + :
Now use Rf = 1= lim supn!1(janj1=n) to nd its radius of convergence.
Hint: you will need to use the result of 3.
6. Find a power series expansion for f(x) = (1+x2)=(1 x) about x = 0 and then nd its radius of convergence by rst expanding 1=(1 x) as a geometric series, multiplying out and then collecting the terms relating to the same power of x. You will need to derive a formula for the n'th coecient.
1st October 2008
1
