Advanced High-School Mathematics

SECTION 5.3 Concept of Power Series 287

∑∞

n=0

n 2 n

2 n

=

∑∞

n=0

n which also diverges.

Therefore,

∑∞

n=0

nxn

2 n

has interval of convergence − 2 < x < 2.

Example 8. The power series

∑∞

n=0

(−1)nxn

n!

has infinite radius of con-

vergence so there are no endpoints to check.

Before closing this section, we should mention that not all power se-

ries are of the form

∑∞

n=0

anxn; they may appear in a “translated format,”

say, one like

∑∞

n=0

an(x−a)n, where ais a constant. For example, con-

sider the series in Example 4, on page 285. What would the interval of
convergence look here? We already saw that this series was guaranteed
to converge on the interval− 4 < x < 0. If x =−4, then this series
is the convergent alternating harmonic series. Ifx= 0, then the series
becomes the divergent harmonic series. Summarizing, the interval of
convergence is− 4 ≤x <0.

Exercises

1. Determine the radius of convergence of each of the power series
   below:

(a)

∑∞

n=0

nxn

n+ 1

(b)

∑∞

n=0

nxn

2 n

(c)

∑∞

n=0

(−1)nx^2 n

n^2 + 1

(d)

∑∞

n=0

(−1)nx^2 n

3 n

(e)

∑∞

n=0

(x+ 2)n

2 n

(f)

∑∞

n=0

(−1)n(x+ 2)n

2 n

(g)

∑∞

n=0

(2x)n

n!

(h)

∑∞

n=1

( xn

1 +^1 n

)n

(i)

∑∞

n=1

nlnnxn

2 n