Consider the set $S$ of sequences $(s_n)$ such that $\sum \limits_{k=1}^{\infty} s_k^2$ converges and define the operations $(s_n)+(t_n) := (s_n+t_n)$ and $\lambda (s_n) := (\lambda s_n)$.
I am trying to show that this is a vector space. It is straightforward to check that $\sum \limits_{k=1}^{\infty} (\lambda s_k)^2$ converges if $\sum \limits_{k=1}^{\infty} s_k^2$ converges. Hence, $\lambda (s_n)$ is in $S$.
But how can I show that $(s_n)+(t_n)$ is in $S$? To do so, I need to show that $\sum \limits_{k=1}^{\infty} (s_k + t_k)^2$ converges.
We know that
$$\sum \limits_{k=1}^{\infty} (s_k + t_k)^2 = \lim \limits_{n \to \infty} \sum \limits_{k=1}^{n} (s_k + t_k)^2 = \lim \limits_{n \to \infty} \sum \limits_{k=1}^{n} s_k^2 + 2s_k t_k + t_k^2 = \lim \limits_{n \to \infty} (\sum \limits_{k=1}^{n} s_k^2 + 2 \sum \limits_{k=1}^{n} s_k t_k + \sum \limits_{k=1}^{n} t_k^2)$$
and so it remains to show that the limit of the middle term exists.
How can I show this?
Thank you very much!
Best Answer
You can use the simple inequality
As is well known, "any absolutely convergent series of complex numbers is convergent". So we examine $$u_n \;=\;\sum \limits_{k=1}^{n} |s_k| |t_k|.$$
Using Lemma 1, with real $C>0$, $$u_n \;\leqslant\;\sum \limits_{k=1}^{n} |s_k| |t_k|\;\leqslant\;C\sum \limits_{k=1}^{n} s_k^2 + C\sum \limits_{k=1}^{n} t_k^2\;\leqslant\;C\sum \limits_{k=1}^\infty s_k^2 + C\sum \limits_{k=1}^\infty t_k^2 < \infty$$ holds for each $n$.
Thus, sequence $u_n$ is bounded above, and is evidently monotone increasing.