Prove that a certain subspace $M$ of space $ (C_{00},\|\cdot\|_2)$ is closed and determine $M^{\perp}$ Could someone help me with this exercise?
Let $$M=\left\{x\in C_{00}: \sum_{k=1}^\infty{\frac{x(k)}{k}=0}\right\}$$ which is a subspace of the prehilbertian space $(C_{00},\|\cdot\|_2)$ . Prove that $M$ is closed and calculate $M^ \perp$.
Clarification
$C_{00}$ is the space of sequences with a finite number of terms different from $0.$

You can see $M$ as the kernel of the linear functional $$\phi:x\longmapsto \sum_k\frac{x(k)}k.$$
So if you show that $\phi$ is bounded, then $M$ is closed. And $\phi$ is bounded because by Cauchy-Schwarz
$$
|\phi(x)|\leq\sum_k\frac{|x(k)|}k\leq \frac{\pi}{\sqrt 6}\,\left(\sum_k|x(k)|^2\right)^{1/2}=\frac{\pi}{\sqrt6}\,\|x\|.
$$
As for $M^\perp$, it will be given by those functionals that have the same kernel as $\phi$. 
Edit: The answer by Fred provides a glimpse into looking at $M^\perp$ from the abstract point of view (and it is what I hinted in my answer). But one can also deal with $M^\perp$ directly. In fact, one can show very easily that $M^\perp=\{0\}$. Indeed, suppose that $y\in M^\perp$, this means that $\sum_k x(k)\overline{y(k)}=0$ for all $x\in M$. Let $m$ be the maximum index such that $y(m)\ne0$. Fix $j\leq m$, and construct an $x\in M$ in the following manner: let $x(j)=1$, $x(m+1)=-(m+1)/j$, and $x(k)=0$ for $k\in\mathbb N\setminus \{j,m+1\}$. 
This way $x\in M$ and $$0=\langle x,y\rangle=\overline{y(j)}.$$
As we can do this for every $j=1,\ldots,m$, it follows that $y=0$. So $M^\perp=\{0\}$. 
