Proving that an infinite Euclidean space is linear vector space Assuming real quantities, with the scalar product defined as
$$\langle x,y \rangle = x_{1}y_{1}+x_{2}y_{2}+\cdots, $$
where $x,y$ are vectors in the real infinite Euclidean space $\mathbb{E}_{\infty}$, and assuming that all vectors in such space are by definition of finite length (e.g. $|x|=\langle x,x \rangle=x^{2}_{1}+x^{2}_{2}+\cdots$ converges to a finite value), how can we prove that the product  $\langle x,y \rangle$ will also exist and be finite?  (In other words, we prove that $\mathbb{E}_{\infty}$ will be a linear vector space.)
I am especially interested if there are two approaches of proof: (1) one that is simple and does not use Cauchy-Schwarz inequality (if possible), and (2) another one that uses the Cauchy-Schwarz inequality. This will help me understand the differences and compare with literature (e.g. Friedman's book, ch-1).
Finally, my attempt for the former was to use the D'Alambert (ratio) test to say that, if $|x|,|y|$ converge, then we know that $\lim_{n \rightarrow \infty} |\frac{x_{n+1}}{x_{n}}|^{2}<1$ and $\lim_{n \rightarrow \infty} |\frac{y_{n+1}}{y_{n}}|^{2}<1$, thus $\lim_{n \rightarrow \infty} |\frac{x_{n+1}}{x_{n}}|<1$ and $\lim_{n \rightarrow \infty} |\frac{y_{n+1}}{y_{n}}|<1$.  Then $\lim_{n \rightarrow \infty} |\frac{x_{n+1}}{x_{n}}\frac{y_{n+1}}{y_{n}}|<1$ gives the absolute convergence of the series for $\langle x,y \rangle$. But I am told this is not a rigorous answer because the convergence of a given series does not tell us anything about the ratio test.  Does this also mean that the ratio test for absolute convergence is sufficient but not necessary?  Any elaboration on this would be appreciated.
 A: The ratio test only says that if $\lim_{n\rightarrow\infty}\left|\frac{a_n}{a_{n+1}}\right| < 1$ then $\sum_{n=1}^{\infty}a_n$ converges absolutely; the converse is false - for instance, $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges despite the fact that the ratio of consecutive terms tends to $1$. Thus, it's not much good here - all you are given is that some sequences converge, but nothing about how quickly they do so.
A better thing to note is that, if you have some sequence of $a_n$, then $\sum_{n=1}^{\infty}a_n$ converges absolutely if and only if there is some upper bound $B$ such that $\sum_{n=1}^N |a_n| \leq B$ for all $N$ - this is, more or less, the monotone convergence theorem for real numbers. This is a more helpful characterization because it is bidirectional - you can use it to transform your givens into some statement about finite sums and then to transform a statement about finite sums back into a statement about limits.
More clearly, your goal could be:

Suppose that there are $B_1$ and $B_2$ such that $\sum_{n=1}^Nx_n^2\leq B_1$ and $\sum_{n=1}^Ny_n^2\leq B_2$ for all $N$. Show that there is some $B$ such that $\sum_{n=1}^N|x_ny_n| \leq B$ for all $N$.

The motivation for transforming the goal thusly is that we have gotten rid of most the analytical difficulties and turned this into an algebra problem.
The most straightforward thing to do would be to say:

Let $B=B_1+B_2$. Note that for any $n$, we have either that $|x_n|\leq |y_n|$ or $|y_n|\leq |x_n|$. In the former case, note $|x_ny_n| \leq y_n^2$ and in the latter $|x_ny_n| \leq x_n^2$. In any case, $|x_ny_n| \leq x_n^2+y_n^2$. Therefore, $\sum_{n=1}^N|x_ny_n| \leq \sum_{n=1}^Nx_n^2 + \sum_{n=1}^N y_n^2 \leq B_1+B_2$ as desired, so $\sum_{n=1}^Nx_ny_n$ converges absolutely.

If you wanted to use Cauchy-Schwarz, you could use it here too:

Let $B=\sqrt{B_1B_2}$. For any $n$, consider the vectors $(|x_1|,\ldots,|x_n|)$ and $(|y_1|,\ldots,|y_n|)$. By the Cauchy-Schwarz inequality applied to these vectors, we have
$$\sum_{n=1}^N |x_ny_n| \leq \sqrt{\sum_{n=1}^N x_n^2 \cdot \sum_{n=1}^N y_n^2} \leq \sqrt{B_1B_2}.$$

This gives the algebraic result you need just as well - and gives a somewhat tighter bound on the ultimate $\sum_{n=1}^{\infty}|x_ny_n|$, although this is a bit redundant since you could prove the Cauchy-Schwarz inequality as soon as you know that this space is in fact an inner product space. Note that the proofs are essentially the same - after transforming the analytical statement into an algebraic one, we just need some algebra to fill in a gap and can do this in a number of ways.
