MATH 106A,B - CALCULUS II WINTER 2009
QUIZ 10
NAME:
Show ALL your work CAREFULLY.
(a) Find the interval of convergence of the following power series [Do not forget to check the endpoints!].
∞
X
n=1
(x−1)n
n3
Applying the ratio test, we have
lim
n→∞
|an+1|
|an|= lim
n→∞
|x−1|n+1
(n+ 1)3.|x−1|n
n3= lim
n→∞
|x−1|n+1
(n+ 1)3·n3
|x−1|n
=|x−1| · lim
n→∞
n3
(n+ 1)3=|x−1| · lim
n→∞ n
n+ 13
=|x−1|.
It follows that the series S(x) = P∞
n=1
(x−1)n
n3converges for |x−1|<1or 0< x < 2. At the
endpoints, we have S(0) and S(2). At x= 1,S(2) = Pn=1 1
n3converges by the p-test. Note that
S(2) is the corresponding series of absolute values for the series S(0) = Pn=1
(−1)n
n3. In other
words, since S(2) converges, S(0) converges absolutely so S(0) converges (or one can use the
Alternating Series Test). Hence, the the interval of convergence for S(x)is [0,2].
(b) Find a power series representation of ln(x+ 1). [What is the derivative of ln(x+ 1)? Recall that
1
1−t=P∞
n=0 tnfor |t|<1.]
Consider the power series representation of 1
1+x, by letting t=−x.
1
1 + x=X
n=0
(−x)n=X
n=0
(−1)nxnfor |x|<1.
It follows that
ln(x+ 1) = Z1
1 + xdx =ZX
n=0
(−1)nxndx
=X
n=0 Z(−1)nxndx for |x|<1
=X
n=0
(−1)nxn+1
n+ 1 =x−x2
2+x3
3−x4
4+... for |x|<1.
In fact, the series above converges when x= 1 but not when x=−1. Thus, we have
ln(x+ 1) = X
n=0
(−1)nxn+1
n+ 1 =x−x2
2+x3
3−x4
4+... for −1< x ≤1.
Bonus: now you know that 1−1
2+1
3−1
4+... = ln 2. Isn’t this COOL?
Date: April 8, 2009.
1