For $x,y\in l_2, x^{(k)}\to x, y^{(k)}\to y$ then $\langle x^{(k)},y^{(k)}\rangle \to \langle x,y\rangle$ 
For $x,y\in l_2$ if $x^{(k)}\to x, y^{(k)}\to y$ then show that $\langle x^{(k)},y^{(k)}\rangle \to \langle x,y\rangle$. Here $x^{(k)} = (x_1,x_2,\ldots,x_k,0,0,\ldots,0)$ and $\langle x,y\rangle = \sum_{i=1}^\infty x_iy_i$.

Just want to check if my proof is alright:
$$x^{(k)}\to x \implies \forall \epsilon > 0 \exists N_1\in\mathbb N\ \forall k\ge N_1 (\|x^{(k)}- x\|_2 < \epsilon)$$
$$y^{(k)}\to y \implies \forall \epsilon > 0 \exists N_2\in\mathbb N\ \forall k\ge N_2 (\|y^{(k)}- y\|_2 < \epsilon)$$
Pick $N_1,N_2$ such that $\|x^{(k)}- x\|_2 < \sqrt\epsilon$ for all $k\ge N_1$ and $\|y^{(k)}- y\|_2 < \sqrt\epsilon$ for all $k\ge N_2$. Put $N=\max (N_1,N_2)$.
Now note that $$|\langle x^{(k)},y^{(k)}\rangle - \langle x,y\rangle| \le \|x^{(k)}- x\|_2\|y^{(k)}- y\|_2 \le \epsilon, \forall k\ge N$$
So, the proof is complete.
Sounds good?
 A: This is a corollary of the following fact:
Let $\mathcal{H}$ be an inner product space, with inner product $\langle\cdot,\cdot\rangle$.
Let $x,y,x_{n},y_{n}\in\mathcal{H}$. If $x_{n}\rightarrow x$ and
$y_{n}\rightarrow y$, then $\langle x_{n},y_{n}\rangle\rightarrow\langle x,y\rangle$.
Proof: Since $(x_{n})$ and $(y_{n})$ are convergent sequences, they
are bounded. Choose $M>0$ such that $||x_{n}||\leq M$ and $||y_{n}||\leq M$
for all $n$. Note that
\begin{eqnarray*}
||x|| & \leq & ||x-x_{n}||+||x_{n}||\\
 & \leq & ||x-x_{n}||+M.
\end{eqnarray*}
Letting $n\rightarrow\infty$ and noticing that $||x-x_{n}||\rightarrow0$,
we have that $||x||\leq M$. Similarly $||y||\leq M$. We also assume
Cauchy-Schwarz ineqality (i.e., $|\langle x,y\rangle|\leq||x||\cdot||y||$)
without proof.
Consider
\begin{eqnarray*}
 &  & \left|\langle x_{n},y_{n}\rangle-\langle x,y\rangle\right|\\
 & \leq & \left|\langle x_{n},y_{n}\rangle-\langle x,y_{n}\rangle\right|+\left|\langle x,y_{n}\rangle-\langle x,y\rangle\right|\\
 & = & \left|\langle x_{n}-x,y_{n}\rangle\right|+\left|\langle x,y_{n}-y\rangle\right|\\
 & \leq & ||x_{n}-x||\cdot||y_{n}||+||x||\cdot||y_{n}-y||\\
 & \leq & M\left(||x_{n}-x||+||y_{n}-y||\right)\\
 & \rightarrow & 0.
\end{eqnarray*}
This shows that $\langle x_{n},y_{n}\rangle\rightarrow\langle x,y\rangle$.
