Calculus Notes
Grinshpan
POWER SERIES
A power series centered at x0 is an infinite sum of the form
a0 + a1 (x − x0 ) + a2 (x − x0 )2 + . . . + an (x − x0 )n + . . . .
A common choice for the center x0 is 0. The numbers a0 , a1 , a2 , . . . are the
coefficients of the power series.
For any given value of x, the powers series becomes a regular infinite series of
numbers, either convergent or divergent. For instance, if x = x0 one always
obtains a convergent sum: a0 + 0 + 0 + 0 + . . . = a0 .
EXAMPLE. The power series
1 + x + x2 + . . . + xn + . . .
is centered at x0 = 0; its coefficients are an = 1, n = 0, 1, 2, . . . . This is our old
friend, the geometric series with ratio x. For every x with |x| < 1 the power series
1
) and for all other x it diverges.
converges (to 1−x
THEOREM. The set of all values of x such that the power series converges is an
interval centered at x0 , called the interval of convergence. The distance R from x0
to either end of this interval is the radius of convergence. Three alternatives are
possible:
R = 0, the power series converges for x = x0 only,
R > 0, the power series converges for |x − x0 | < R and diverges for |x − x0 | > R,
R = ∞, the power series converges for all x without restriction.
For every x with |x − x0 | < R, the power series converges absolutely. If R is finite
and positive, the endpoints x = x0 ± R must be considered separately: they may
or may not be points of convergence. For x varying in the interval of convergence,
the power series represents a function of x, namely, its sum.
To understand the logic behind this theorem consider the next example.
EXAMPLE. The power series
x+
1
2
x2 +
1
3
x3 + . . . +
1
n
xn + . . .
is centered at x0 = 0, its coefficients are an = n1 , n = 1, 2, 3, . . . . Fix x and apply
the Ratio test:
|x|n+1 |x|n
n
:
=
|x| → |x|, n → ∞.
n+1
n
n+1
Hence for every x with |x| < 1 the power series converges and, in fact, absolutely.
For every x with |x| > 1 the power series diverges. For x = ±1 the Ratio test is
inconclusive, these cases need separate consideration (endpoint analysis).
2
If x = 1, the power series becomes harmonic:
1+
1
2
+
1
3
+
1
4
+ ... .
Divergent. If x = −1, the resulting series
−1 +
1
2
−
1
3
+
1
4
− ...
converges by the alternating series test.
It follows that the interval of convergence is −1 ≤ x < 1.
1
In this interval the sum of the power series can be shown to be ln( 1−x
).
THEOREM. The radius of convergence of the power series can be found by either
formula
1
|an |
R = lim p
,
R = lim
,
n→∞ n |a |
n→∞ |an+1 |
n
provided that the limit in question exists.
EXAMPLE. The power series
∞
X
xn
n!
n=0
is centered at x0 = 0; its coefficients are an =
1
n! ,
n = 0, 1, 2, . . . . Observe that
|an |
(n + 1)!
= lim
= lim (n + 1) = ∞.
n→∞
|an+1 | n→∞
n!
Hence the radius of convergence is infinite. This means that the power series
converges for all values of x. The function represented by this power series is ex .
lim
n→∞