Reflexivity of $\ell^p$

So you have the canonical maps:

$$\begin{cases} f_q: \ell^p\to (\ell^q)^* \\ f_p: (\ell^q)^*\to (\ell^p)^{**}\end{cases}$$

which are the isomorphisms between $\ell^p$ and $(\ell^q)^*$ and $\ell^q$ and $(\ell^p)^*$ respectively. Then just write $f_p\circ f_q=j_p:\ell^p\to (\ell^p)^{**}$.

This is the answer I've unravelled:

Let $z \in (\ell^p)^{\ast\ast}$. I want to prove that exists $x \in \ell^p$ such that $\langle z,f\rangle=\langle f,x \rangle$ for every $f \in (\ell^p)^\ast$.

I know that there are the isomorphisms:

$j_p: \ell^q \rightarrow (\ell^p)^\ast$ and $j_q: \ell^p \rightarrow (\ell^q)^\ast$

Now, fix $z \in (\ell^p)^{\ast\ast}$.

$\langle z, f\rangle = \langle z, j_p(y)\rangle$ for some $y$ in $\ell^q$.

I define $g(y)=\langle z, j_p(y)\rangle$. It's seen that $g$ is an element of $(\ell^q)^\ast$ then

$\langle z, f\rangle = \langle z, j_p(y)\rangle = \langle g, y \rangle$

Being an element of $(\ell^q)^\ast$, the number $\langle g, y \rangle$ can be represented as $\sum_{k=1}^\infty x_ky_k$ for some $x \in \ell^p$

Now fix that $x$. This sum, $\sum_{k=1}^\infty x_ky_k$ can be seen as a functional $u$ on $\ell^p$ described by $j_p(y)$.

Then $u=f$ and $\langle z, f\rangle = \langle z, j_p(y)\rangle = \langle g, y \rangle = \sum_{k=1}^\infty x_ky_k = \langle f,x \rangle$ as I wanted.

${}$

Now the answer seems correct to me, but still a bit confused because I would like to avoid to mix the $\langle \cdot, \cdot \rangle$ with the sum notation. And some suggestions to avoid that are appreciated.

After I would like to understand why the answer provided by Adam Hughes is rigorous: why does he claims that his $f_p: (\ell^q)^*\to (\ell^p)^{**}$ is the equality isomorphism? Indeed it's needed the isomorphism $j_p: \ell^q \rightarrow (\ell^p)^\ast$ to show that it is the equality isomorphism, but how? And then, acknowledged that his $f_p\circ f_q$ is an isomorphism from $l^p$ to $(l^p)^{\ast\ast}$, why it is said that it's equal to the canonical isomorphism?

I'm slow in understanding... I know! :-p