The norm of an element $(\gamma_k)=x \in l^{\infty}$ is defined to be $\sup_k \gamma_k <\infty$. The norm of an element $(\gamma_k)=x \in l^p$ is defined to be $(\sum_k \gamma_k^p)^{1/p} < \infty$, even for $p=1$!
I can't see how $\displaystyle{\lim_{p \to \infty}} (\sum_k \gamma_k^p)^{1/p} = \sup_k \gamma_k$ holds?
And, if they just are separate definitions, then why the case '$p=\infty$' is in the 'category' of $L^p$ spaces?
Added - Understanding answers to this question requires a knowledge more than an undergraduate Analysis, i.e. my current level. Simple detailed answers would be much appreciated.
In general if $f \in L^p$ for some $1 \leq p<\infty$ and $f \in L^\infty$ then $f \in L^q$ for all $q \in [p,\infty]$. In this case $\| f \|_{L^\infty} = \lim_{p \to \infty} \| f \|_{L^p}$. This is easiest to see in the case of $\mathbb{R}^n$, where (assuming $x_i \neq 0$)
$$\| x \|_p = |x_i| \left ( \sum_{j=1}^n \left ( \frac{|x_j|}{|x_i|} \right )^p \right )^{1/p}$$
If $|x_i|$ is the biggest, then all the terms either go to zero or $1$. In either case since $n$ is fixed,
$$\left ( \sum_{j=1}^n \left ( \frac{|x_j|}{|x_i|} \right )^p \right )^{1/p} \to 1$$
as $p \to \infty$. The idea in the general case is that the condition $f \in L^p$ ensures that $\left ( \frac{|f(x)|}{\| f \|_{L^\infty}} \right )^p$ goes to zero rapidly with $p$ for "most" $x$.