Item 1 in your definition of an inner product is incorrect. A simple counter example from $\mathbb{R}^2$ is
$$(1,0).(-1,0) = -1.$$
It should read
$$\langle v, v \rangle \ge 0,$$
this guarantees that all vectors in your space have a non-negative length.
It's clear that if at least one of $\|u\|,\|v\|,\|w\|$ is $0$ the inequality is true. Therefore we may assume without loss of generality that $\|u\|,\|v\|,\|w\|$ are all non-zero.
Define $u'=\dfrac{u}{\|u\|}$, $v'=\dfrac{v}{\|v\|}$ and $w'=\dfrac{w}{\|w\|}$. Dividing both sides by $\|u\|^2\|v\|\|w\|$ we get that the inequality is true for $u,v,w$ if and only if it is true for $u',v',w'$, therefore, we may assume that $\|u\|=\|v\|=\|w\|=1$. Then we need to prove that $2\langle u,v\rangle \langle u,w\rangle \leq 1+\langle v,w\rangle$.
Now, if $\lambda \in \mathbb{R}$, we have\begin{align*}0 & \leq \|\lambda u+v+w\|^2 \\
& =\lambda ^2\|u\|^2+2\lambda (\langle u,v\rangle +\langle u,w\rangle )+(\|v\|^2+\|w\|^2+2\langle v,w\rangle ) \\
& =\lambda ^2+2(\langle u,v\rangle +\langle u,w\rangle )+2(1+\langle v,w\rangle ).
\end{align*}This is a quadratic equation in $\lambda$ that has at most one real root, therefore, its discriminant $\Delta$ satisfies $\Delta \leq 0$. This implies that$$(\langle u,v\rangle +\langle u,w\rangle )^2\leq 2(1+\langle v,w\rangle ).$$Now, if $a,b\in \mathbb{R}$, we have that $0\leq (a-b)^2=a^2-2ab+b^2$, which implies that $4ab\leq a^2+2ab+b^2=(a+b)^2$. In particular, with $a=\langle u,v\rangle$ and $b=\langle u,w\rangle$, we get$$4\langle u,v\rangle \langle u,w\rangle \leq (\langle u,v\rangle +\langle u,w\rangle )^2\leq 2(1+\langle v,w\rangle ).$$Therefore,$$2\langle u,v\rangle \langle u,w\rangle \leq 1+\langle v,w\rangle$$as wanted.
Best Answer
$\sum\limits_{n=1}^{N}|u_nv_n| \leq \sqrt {\sum\limits_{n=1}^{N}|u_n|^{2}} \sqrt {\sum\limits_{n=1}^{N}|u_n|^{2}} $ for all $N$ (by Cauchy-Schwarz inequality in $\mathbb C^{N}$). Since the right side is bounded it follows that the series $\sum u_nv_n$ is absolutely convergent.