Demonstrate the following sequence is Cauchy sequence...

Question

Here I have such an exercise:

Show that the sequence $(x_n)$,where

is a Cauchy sequence.

Here's the solution:

Can you, please, take a look and explain me why the relation marked with $"?"$ is true, and why is it necessary? Thank you very much.

Answer 1

$$x_{n+p}-x_n=\sum_{k=0}^{n+p} a_k-\sum_{k=0}^na_k=$$

$$=a_0+a_1+\ldots+a_n+\ldots+a_{n+p}-(a_0+a_1+\ldots+a_n)=a_{n+1}+a_{n+2}+\ldots+a_{n+p}$$

It's necessary to do this as this is Cauchy's condition.

Answer 2

Let us write $C_k$ for the members $\displaystyle\frac{\cos(k\alpha)}{2^k}$, then we have $$\begin{align} x_n &=C_0+C_1+\dots+C_n \\ x_{n+p} &= C_0+C_1+\dots+C_n+C_{n+1}+\dots+C_{n+p} \end{align}\,.$$ Thus, $x_{n+p}-x_n=C_{n+1}+\dots+C_{n+p}\ $ (and actually we get strict equation instead of $\le$).