282 CHAPTER 5 Series and Differential Equations
Sincebk → 0 ask → ∞, we conclude that (rn) is a Cauchy sequence
of real numbers, hence converges.
As a corollary to the above, let’s consider the convergence of the
sequence alluded to above, viz.,
∑∞
n=1
cosn
n
. To do this, we start with the
fact that for all integersk,
2 sin(1/2) cosk= sin(k+ 1/2)−sin(k− 1 /2).
Therefore,
∣∣
∣2 sin(1/2)
∣∣
∣∣
∣·
∣∣
∣∣
∣
∑n
k=1
cosk
∣∣
∣∣
∣ =
∣∣
∣∣
∣
∑n
k=1
(sin(k+ 1/2)−sin(k− 1 /2))
∣∣
∣∣
∣
=
∣∣
∣∣
∣sin(n+ 1/2)−sin(1/2)
∣∣
∣∣
∣
≤ 2.
Since sin(1/2) 6 = 0 we already see that
∑n
k=1
coskis bounded, and Dirich-
let’s theorem applies, proving the convergence of
∑∞
n=1
cosn
n
Exercises
- Strenghten the result proved above by proving that the series
∑∞
n=1
cosnx
np
converges whenever p > 0, and x is not an integral
multiple of 2π.
- Prove that
∑∞
n=1
sinnx
np
converges wheneverp >0.
5.3 The Concept of a Power Series
Let us return to the familiar geometric series, with ratior satisfying
|r|<1.