4.1 • SEQUENCES AND SERIES 1 3 1- EXAMPLE 4.6 Show that f; %"!:~1~ converges.
n = l
Solution We calculate lznl =I <^35 "!;~ ;· 1 = ,& = M.,... Using the comparison test
and the fact that n~l ;, converges, we determine that n~l I %"!:~J~ I converges
and hence so does f; %~~J~.
n=I
-------~EXERCISES FOR SECTION 4.1- Find the following limits.
(a) n -Jim oo (! + t)".
(b) Jim n+(i)".
n-oo n
( ) r n(^2) ±i2"
e n~ 2n.
( d) Jim (n+i)<j+ni).
n-oo n
I I
- Show that Jim (i)" = 1, where (i)" is the principal value of the nth root of i.
n-oo - Suppose that Jim Zn = zo. Show that lim z;;-= zo.
n-oo n -oo - Suppose that the complex series {z,.} converges to(. Show that {Zn} is bounded
in two ways.
(a) Write Zn = x.,.. + iy,., and use the fact that convergent series of real
numbers are bounded.
(b) For e = 1, use Definitions 4.1 and 4.2 to show that there is someinteger N such that, for n > N, IZnl = I(+ (z,. -()I :::; 1(1+1. Then
set R = max{lz1l .lz2I , ... lzNI ,( + l}.
5. Show that f: (n+~+i - n~i) = i.
·n=O
00 00 -- Suppose that E Zn= S. Show that E z;;-= S.
n = l ns=l - Does .. ~~·(.!Jrr exist? Why?
- Let Zn = r n e'^9 n :/; 0, where e,, = Arg (Zn).
(a) Suppose Jim Tn = ro and Jim e .. = 80. Show Jim r.,..e•^9 n = roe'^90.
n -oo n -oo n -oo(b) Find an example where Jim Zn = Zo = r 0 ei^90 , Jim r,. = r 0 , but
n- oo n-ooLim e .. does not exist.
n-oo
(c) Is it possible to have n-oo lim Zn = Zo = r 0 e'^9 o, but n-oo Jim r,. does not exist?