But I wouldn't sit down and read it straight through to begin the book. The homotopy extension stuff in particular will seem super boring and you won't need it for a while, if ever.
Once you made that split you did in the comment we may go this way: 1 + $\lim \limits_{n \to \infty} \left( \dfrac{n-1}{n} \right)^n$ + $\lim \limits_{n \to \infty} \left( \dfrac{n-2}{n} \right)^n$+ $\lim \limits_{n \to \infty} \left( \dfrac{n-3}{n} \right)^n$ + ...
It depends on the case. Here as you notice all terms are different from 0. It's not the case 0*oo as it happens when you have to calculate 1/(n+1) + 1/(n+2) + ...+1/2n
@PeterTamaroff: for my last example, it would be a mistake to calculate separately each limit and then to add them up and say it's 0.
@PeterTamaroff Your squeezings give you an error that must be controlled. For each $n$, there are $2^{n-1}-1$ terms that you are adding, with errors of $\frac{z}{2^{n-1}}$. That has to be controlled.
@PeterTamaroff The Squeeze theorem works for a single limit, but you are adding a lot of them. The number of terms goes up as your squeezing gets tighter.
For each $n$, say we know that $$ \frac{1}{k^2}-\frac1n\le a_{n,k}\le\frac{1}{k^2} $$ So that $\displaystyle\lim_{n\to\infty}a_{n,k}=\frac{1}{k^2}$. Can we say that $$ \lim_{n\to\infty}\sum_{k=1}^na_{n,k}=\frac{\pi^2}{6}\text{?} $$
According to what I've written I have that $$\left| {\frac{z}{{{2^n}}}\left( {\cot \frac{{z + k\pi }}{{{2^n}}} + \cot \frac{{z - k\pi }}{{{2^n}}}} \right) - \frac{{2{z^2}}}{{{z^2} - {k^2}{\pi ^2}}}} \right| < \frac{{\left| z \right|}}{{{2^{n - 1}}}}$$
