Iterated duals of Banach spaces Let $B$ be a (non-reflexive) Banach space. Denote by $B^{(n)}$ the $n$-fold dual, i.e. $B^{(0)}=B$ and $B^{(n)}=(B^{(n-1)})^\prime$. Does there exist an $n\ge 1$ s.t.
$B^{(n)}\cong  B$ ? Please explain.
 A: It may happen that $B^{(2)}\cong B$, even if $B$ is not reflexive, as James' space shows, see James, A non reflexive Banach space isometric with its second conjugate space, Proc. Nat. Acad. Sc. USA, 37 (1951), p.174-177) for instance. 
But it's possible such a $n$ doesn't exists.

Lemma: If $E$ is a Banach space and $E'$ its topological dual space, and $E'$ is separable, so is $E$. 

If $\{f_n\}$ is dense in the unit ball of $E'$, for each $n$, fix 
$x_n\in E$ of norm $1$ such that $|f_n(x_n)| \geq \frac 12\lVert f_n\rVert_{B'}$. 
We have to show that $F:=\operatorname{Span}\{x_k,k\geq 1\}$ is dense in $E$. Let $f$ a continuous linear functional with vanishes on $F$. We have to show that it vanished on $E$. Let $\{n_k\}$ such that $\lVert f-f_{n_k}\rVert\leq k^{-1}$ (we can assume that $f$ has a norm $\leq 1$). Then 
$$\lVert f_{n_k}\rVert\leq \lVert f_{n_k}(x_{n_k})-f(x_{n_k})\rVert\leq k^{-1},$$
so $\lVert f\rVert\leq 2k^{-1}$.
To apply this to our problem, take $B$ a separable Banach space with non-separable dual (as $\ell^1(\Bbb R)$). It's not reflexive, otherwise $B''$ would be separable and so would be $B'$. If $B^{(n)}\cong B$ for some $n\geq 2$, then $B^{(n)}$
 would be separable, as this property is conserved by isomorphism. By the lemma $B^{(n-1)}$ would be separable, and iterating that $B'$^would be separable, a contradiction.
