4.5 POWER SERIES
Determine the range of values ofxfor which the following power series converges:
P(x)=1+2x+4x^2 +8x^3 +···.By using the interval-of-convergence method discussed above,
ρ= lim
n→∞∣
∣∣
∣
2 n+1
2 nx∣
∣∣
∣=|^2 x|,and hence the power series will converge for|x|< 1 /2. Examining the end-points of the
interval separately, we find
P(1/2)=1+1+1+···,
P(− 1 /2) = 1−1+1−···.ObviouslyP(1/2) diverges, whileP(− 1 /2) oscillates. ThereforeP(x) is not convergent at
either end-point of the region but is convergent for− 1 <x<1.
The convergence of power series may be extended to the case where theparameterzis complex. For the power series
P(z)=a 0 +a 1 z+a 2 z^2 +···,we find thatP(z) converges if
ρ= lim
n→∞∣
∣
∣
∣an+1
anz∣
∣
∣
∣=|z|nlim→∞∣
∣
∣
∣an+1
an∣
∣
∣
∣<^1.We therefore have a range in|z|for whichP(z) converges, i.e.P(z)converges
for values ofzlying within a circle in the Argand diagram (in this case centred
on the origin of the Argand diagram). The radius of the circle is called the
radius of convergence:ifzlies inside the circle, the series will converge whereas
ifzlies outside the circle, the series will diverge; if, though,zlies on the circle
then the convergence must be tested using another method. Clearly the radius of
convergenceRis given by 1/R= limn→∞|an+1/an|.
Determine the range of values ofzfor which the following complex power series converges:P(z)=1−z
2+
z^2
4−
z^3
8+···.
We find thatρ=|z/ 2 |, which shows thatP(z)convergesfor|z|<2. Therefore the circle
of convergence in the Argand diagram is centredon the origin and has a radiusR=2.
On this circle we must test the convergence by substituting the value ofzintoP(z)and
considering the resulting series. On the circle of convergence we can writez=2expiθ.
Substituting this intoP(z), we obtain
P(z)=1−2expiθ
2+
4exp2iθ
4−···
=1−expiθ+[expiθ]^2 −···,which is a complex infinite geometric series with first terma= 1 and common ratio