The space of sequences which are eventually zero in $l^2$ is not a Hilbert space. Define $V$ to be the space of sequences which are eventually zero, i.e. 

$$V=\bigcup_{N=1}^{\infty}\{(x_n)_{n=1}^{\infty}\in l^2: x_n=0 \; \text{for}\; n\ge N\}.$$

Is $V$ a Hilbert space with respect to the $l^2$ inner product?

I don't think this is a Hilbert space. The example I came up with is $x^n=(1,1/2,\dots,1/n,0,\dots)$. Then $x^n\in V$, and $||x^n-x^m||_2\to 0$ as $m,n \to \infty$, by the tail convergence as $\sum 1/n^2$ converges. Also, this sequence clearly converges to $(1,1/2,1/3,\dots, 1/n, 1/(n+1),\dots)$, by the same reasoning as above, yet this element is not in $V$. 

This is the simplest example I could come up with, but what other examples aside from the harmonic series can we think of? I would appreciate any suggestions.
This was recently discussed in my Real Analysis class, so this is not my idea. 

First, we need to state the following result. 

If $H$ is a Hilbert space (complete, inner product space), and $M$ is a closed proper subspace of $H$, then there exists a $z \in H, z \neq \vec{0} $ such that $z\bot M$. This result will fail to hold if $H$ is not complete (not Hilbert).

We shall define $V$ as in the question and construct a closed proper subspace $M$ of $V$ and show that if $z \bot M$ then $z = \vec{0}$.

$$V:=\{x:\mathbb{N}\to\mathbb{R}: \exists n_{0} \in \mathbb{N}: x_{n} = 0 \ \forall\ n\geq n_{0} \}$$
Let us endow $V$ with the inner product $$&lt;x,y&gt; = \sum_{n=1}^{\infty}x_{n}y_{n}$$ 
Let $f:V\to \mathbb{R}$ with
$$ f(x) = \sum_{n=1}^{\infty}\frac{x_{n}}{n}$$
We can check that $f$ is linear and $f$ is bounded. 
$$ |f(x)| = |\sum_{n=1}^{\infty}\frac{x_{n}}{n}| \leq \bigg( \sum_{n=1}^{\infty}\frac{1}{n^{2}} \bigg)^{1/2}\bigg( \sum_{n=1}^{\infty}x_n^{2} \bigg)^{1/2}=\sqrt{\frac{\pi^{2}}{4}}||x|| $$

Let us now define the kernel of $f$ as $M:=\{x \in V:f(x)=0\}.$ Then $M$ is a closed linear subspace of $V$ and also a proper subset of $V$.

We will now show that $z = \vec{0}$ is the only element of $V$ such that $z\bot M$, implying that $V$ is not complete, hence not Hilbert.

Consider $z = (z_{n})_{n} \in V$, and $y_{n} = e_{1} - n e_{n}$, where the $i^{th}$ entry of $e_{i}$ is 1, and rest all are $0$.We can observe that $y_{n} \in M$. 
$$ f(y_{n}) = 1 + \bigg(\frac{-n}{n}\bigg)=0 $$
If $z\bot M$, then $z \bot y_{n}, i.e\ &lt;z,y_{n}&gt;=0\ \forall n \in \mathbb{N}$. 
$$ &lt;z,y_{n}&gt; = z_{1} - n z_{n} = 0 \implies z_{n} = \frac{z_{1}}{n}.$$

Since $z \in V, \exists\ n_{0}: z_{n_{0}}=0 \implies z_{1}=0$.

If $z \bot M$, we showed that $z = \vec{0}$, hence $V$ is not complete. So $V$ is not Hilbert.
If an = 2-n/2 for n >= 1, define a  =  (an), a point in el2, N0, (N0 = {0, ..., infinity}), belongs to the unit sphere S.

If we like, we can also divide the finite-dimensional approximations, say bn,k = ak if 0 &lt;= k &lt;=n and 0 if k > n, by their own lengths to also get a sequence of points, say cn, on the unit sphere S of el2(N). Of course these points approach the point a in the el2 limit.

The only advantage being: Now, the points of the eventually-zero ones whose el2 limit is not another eventually-zero point aren't just anywhere, they're on the unit sphere.