Let $f_n:[-1,1]\to\mathbb R$ be such that $$f_n(t)=\begin{cases}1, & \text{if $t\in[-1,0];$} \\1-nt, & \text{if $t\in[0,\tfrac1n]$;} \\ 0, & \text{otherwise.}\end{cases}$$
According to Mathematica, we have $\lVert f_n-f_m\rVert=\frac{(m-n)^2}{3 m^2 n}$ if $1<n<m$ so this is indeed a Cauchy sequence.
In[1]:= f[n_] := Piecewise[{{1, t < 0}, {1 - n t, 0 <= t <= 1/n}}];
In[2]:= Integrate[(f[n]-f[m])^2, {t, -1, 1}, Assumptions-> 1<n<m]
2
(m - n)
Out[2]= --------
2
3 m n
Can you show it does not converge?
The map $\phi : H \to H'$ given by $\phi(v) = f_v$, where $f_v(x) = \langle x, v\rangle$, for $x \in H$ is an antilinear bijective isometry. Bijectivity and norm-preservation both follow from the Riesz Representation Theorem, and antilinearity can be easily verified:
$$\phi(\alpha v + \beta w)(x) = f_{\alpha v + \beta w}(x) = \langle x, \alpha v + \beta w \rangle = \overline{\alpha}\langle x, v\rangle + \overline{\beta}\langle x, w\rangle = \overline{\alpha}f_v(x) + \overline{\beta}f_w(x) = \big(\overline{\alpha}\phi(v) + \overline{\beta}\phi(w)\big)(x)$$
Hence, $\phi(\alpha v + \beta w) = \overline{\alpha}\phi(v) + \overline{\beta}\phi(w)$.
Now, as you can show, if two spaces are "antilinearly isometric", then one is complete if and only if the other one is complete.
Indeed, let $(f_n)_{n=1}^\infty$ be a Cauchy sequence in $H$. We have $f_n = \phi(x_n)$ for some $x_n \in H$.
The sequence $(x_n)_{n=1}^\infty$ is a Cauchy sequence in $H$:
$$\|x_m - x_n\| = \|\phi(x_m - x_n)\| = \|\phi(x_m) - \phi(x_n)\| = \|f_m -
f_n\| \xrightarrow{m, n \to\infty} 0$$
Since $H$ is complete, $(x_n)_{n=1}^\infty$ converges. Set $x_n \xrightarrow{n\to\infty} x \in H$.
We claim that $f_n \xrightarrow{n\to\infty} \phi(x)$ in $H'$. Indeed:
$$\|\phi(x) - f_n\| = \|\phi(x) - \phi(x_n)\| = \|\phi(x - x_n)\| = \|x - x_n\| \xrightarrow{n\to\infty} 0$$
Hence, $H'$ is complete.
Best Answer
In this answer, I will use $x_n$ as a sequence in $l^2$ and write $x_n(k)$ as the $k$-th member of that sequence.
The norm in the Hilbert space is given by $\|x\| = \sqrt{\langle x, x \rangle}$. We wish to show that if a sequence $\{ x_n \} \subset l^2$ is Cauchy, then it converges in $l^2$.
Suppose that $\{x_n\}$ is such a Cauchy sequence. Let $\{ e_k \}$ be the collection of sequences for which $e_k(i) = 1$ if $i=k$ and zero if $i\neq k$.
Then $\langle x_n, e_k \rangle = x_n(k)$. Notice that $$|x_n(k) - x_m(k)| = |\langle x_n - x_m, e_k \rangle| \le \|x_n-x_m\| \| e_k\| = \|x_n-x_m\|$$ for all $k$ (also note that this convergence is uniform over $k$). Therefore the sequence of real numbers given by $\{x_n(k)\}_{n\in \mathbb{N}}$ is Cauchy for each $k$, and thus converges. Call the limit of this sequence $\tilde x(k)$.
Let $\tilde x = (\tilde x(k))_{k\in\mathbb{N}}$. We wish to show that $\tilde x \in l^2$.
Consider $$\sum_{k=1}^\infty |\tilde x(k)|^2=\sum_{k=1}^\infty |\lim_{n\to\infty} x_n(k)|^2=\lim_{n\to\infty} \sum_{k=1}^\infty |x_n(k)|^2=\lim_{n\to\infty}\|x_n\|^2.$$
The exchange of limits is justified, since the convergence of $\lim_{n\to\infty} x_n(k)$ is uniform over $k$. Finally, since $\{ x_n \}$ is Cauchy, the inequality, $$| \|x_m\| - \|x_n\| | < \| x_m - x_n\|$$ implies that $\|x_n\|$ is a Cauchy sequence of real numbers, and so $\|x_n\|$ converges. Thus $\tilde x$ is in $l^2$.
Edit: Completing the proof as per the comments.
We have thus shown that $\tilde x$ is in $l^2$. $\tilde x$ is the most likely candidate for the Cauchy sequence to converge to, and it has been demonstrated to be in our space. What remains is to show that $$\| x_n - \tilde x\| \to 0$$ as $n \to \infty$.
We will utilize a generalized form of the dominated convergence theorem for series. This states that if $a_{n,k} \to b_k$ for all $k$, $a_{n,k} < d_{n,k}$ and $\sum_{k} d_{n,k} \to \sum_{k} D_k < \infty$, then $\lim_{n \to \infty} \sum_{k=0}^\infty a_{n,k} = \sum_{k=0}^\infty b_k$. (here $a_{n,k}, b_k, d_{n,k}, D_{k}$ are all real numbers)
Writing $$\| x_n - \tilde x\|^2 = \sum_{k=0}^\infty |x_n(k) - \tilde x(k)|^2.$$
We see that in this case $a_{n,k} = |x_{n}(k) - \tilde x(k)|^2$, $b_k = 0$, and we must find a $d_{n,k}$ that "dominates" $a_{n,k}$ to finish the proof.
Now note that $|x_n(k) - \tilde x(k)|^2 \le 2 |x_n(k)|^2 + 2 |\tilde x(k)|^2$ and $$\lim_{n \to \infty} \sum_{n=0}^\infty ( 2 |x_n(k)|^2 + 2 |\tilde x(k)|^2) = \sum_{k=0}^\infty (2 |\tilde x(k)|^2 + 2 | \tilde x(k)|^2).$$ Recall that we demonstrated $\lim_{n \to \infty} \sum_{n=0}^\infty |x_n(k)|^2 = \sum_{n=0}^\infty |\tilde x(k)|^2$ in the first half. Thus $D_k$ is played by $4|\tilde x(k)|^2$ in this case.
Thus by the dominated convergence theorem we may conclude that $$\sum_{k=0}^\infty |x_n(k)-\tilde x(k)|^2 \to 0.$$