Convergence of Square of Linear Combination of Sequences whose Squares Converge

Theorem
Let $\left\langle{x_i}\right\rangle$ and $\left\langle{y_i}\right\rangle$ be real sequences such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ and $\displaystyle \sum_{i \mathop \ge 0} y_i^2$ are convergent.

Let $\lambda, \mu \in \R$ be real numbers.

Then $\displaystyle \sum_{i \mathop \ge 0} \left({\lambda x_i + \mu y_i}\right)^2$ is convergent.

Proof
Let $n \in \N$.

Then:
 * $\displaystyle \sum_{i \mathop = 1}^n \left({\lambda x_i + \mu y_i}\right)^2 = \lambda^2 \sum_{i \mathop = 1}^n x_i^2 + \mu^2 \sum_{i \mathop = 1}^n y_i + 2 \lambda \mu \sum_{i \mathop = 1}^n x_i y_i$

By Cauchy's Inequality:
 * $\displaystyle \sum_{i \mathop = 1}^n x_i y_i \le \left({\sum_{i \mathop = 1}^n x_i^2}\right)^{\frac 1 2} \left({\sum_{i \mathop = 1}^n y_i^2}\right)^{\frac 1 2}$

Hence:

Thus the sequence of partial sums $\displaystyle \sum_{i \mathop = 1}^n \left({\lambda x_i + \mu y_i}\right)^2$ is bounded above.

We also have that $\displaystyle \sum_{i \mathop = 1}^n \left({\lambda x_i + \mu y_i}\right)^2$ is also increasing.

So by the Monotone Convergence Theorem, $\displaystyle \sum_{i \mathop = 1}^n \left({\lambda x_i + \mu y_i}\right)^2$ is convergent.