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.
 A: A general (yet vague) principle: on any reasonable space that "occurs in nature" (i.e., is not some artificial contraption concocted for something to do), there is only one topology that one should use and which is natural for this space. Moreover, if you don't use this topology, bad things happen.
$V$ with $||\cdot||_2$ is a well defined pre-Hilbert space which is a normed space, but it is not a Hilbert space because you don't have completeness. This is an example of bad thing happening because you did not pick the natural topology of $V$. The latter is the finest locally convex topology, namely the locally convex topology defined by the collection of all seminorms on $V$.
A: 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 $$<x,y> = \sum_{n=1}^{\infty}x_{n}y_{n}$$ 
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 $z \bot y_{n}, i.e\ <z,y_{n}>=0\ \forall n \in \mathbb{N}$. 
$$ <z,y_{n}> = z_{1} - n z_{n} = 0 \implies z_{n} = \frac{z_{1}}{n}.$$
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.
A: The following example resembles yours, with the harmonic series, but is an exponential series:
For each positive integer $k$, the element
$$
a_{k} = \left( {1 \over 2}, {1 \over 2^2}, {1 \over 2^3}, \ldots, {1 \over 2^{k}}, 0, 0, \ldots \right)
$$
is in $V$.  The sequence $\{ a_{k} \}_{k \geq 1}$ converges in $l^2$ to the element
$$
a = \left( {1 \over 2}, {1 \over 2^2}, {1 \over 2^3}, \ldots, \right),
$$
whose $n$-th term is $2^{-n}$.  Now, $a$ is in $l^2$, but not in $V$.
Another example is produced by replacing with zero, say, every even-positioned element in $a$ and in every $a_{k}$.
In general, if you take any absolutely convergent series
$$
\sum_{n \geq 1} x_{n},
$$
then the element
$$
\left( \sqrt{|x_{1}|}, \sqrt{|x_{2}|}, \ldots, \right)
$$
will be in $x$ an can be approximated by its projections onto the first $k$ coordinates, with $k = 1, 2, 3, \ldots$.
Notice that, by hook or by crook, the limiting element of $l^2$ must have infinitely many nonzero terms for us to "get out of" $V$.
A: All the answers are very nice, but I just want to add the fact that -: "Every infinite dimensional Banach space can't have countable basis." (Baire Category Theorem)
Answer of your question is not too far from  here.
A: 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 <= k <=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.
