Advanced High-School Mathematics

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.