MATH 106B,C - CALCULUS II FALL 2011
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!].
โ
X
n=1
2n
3n(x+ 3)n
Applying the Ratio Test to the corresponding series of absolute values, we have
L= lim
nโโ
2n+1
3(n+ 1)|x+ 3|n+1/2n
3(n)|x+ 3|n
= lim
nโโ 2n
n+ 1 ยท |x+ 3|= 2|x+ 3|.
If 2|x+ 3|<1then the original power series converges and it may converge at the endpoints.
Note that 2|x+ 3|<1is equivalent to |x+ 3|<1
2which in turn is equivalent to โ7
2<x<โ5
2.
At x=โ5
2, the series is Pโ
n=1 2n
3n๎1
2๎n=Pโ
n=1 1
3nand it diverges since it is one-third of the
harmonic series.
At x=โ7
2, the series is Pโ
n=1 2n
3n๎โ1
2๎n=Pโ
n=1
(โ1)n
3nand it converges since it is negative one-
third of the alternating harmonic series.
Thus, we conclude that the interval of convergence is โ7
2โคx < 5
2.
(b) Find a power series representation of sin x
x. State the interval of convergence.
Recall that
sin x=xโx3
3! +x5
5! โx7
7! +... for all x.
If x6= 0 then
sin x
x= 1 โx2
3! +x4
5! โx6
7! +... for all xexcept x= 0.
Hence, the interval of convergence is โโ <x<0โช0<x<โ.
Date: December 7, 2011.
1