As Gerry mentioned, in the first line of (ii) what should have been written is $ S_n = \sum_{k=0}^n {2^{k-1}\over (2^{k }+1)-(2^{k-1}+1)}. $
What's being done in (ii) is the author shows that the infinite sum $\sum\limits_{k=0}^\infty {2^{k-1}\over (2^{k }+1)-(2^{k-1}+1)}$ converges to $L$ (the value of which is found at the end) by showing that the sequence of partials sums $(S_n)$ defined by $S_n=\sum\limits_{k=0}^n {2^{k-1}\over (2^{k }+1)-(2^{k-1}+1)}$ converge to $L$: $\tag{1} \sum\limits_{k=0}^\infty {2^{k-1}\over (2^{k }+1)-(2^{k-1}+1)}=\lim_{n\rightarrow\infty} S_n =\lim_{n\rightarrow\infty} \sum\limits_{k=0}^n {2^{k-1}\over (2^{k }+1)-(2^{k-1}+1)}. $
So in the first line of (ii) (with the corrections that the upper limit in the sums are $n$), the author explicitly finds the value of $S_n$ (which is a finite sum) by using the cancellation "trick" mentioned in the other answers.
He finds, as illustrated by Anon, $S_n={1\over 2^{0-1}+1}-{1\over 2^n -1}$.
That's the value of $S_n$, the sum of the first $n+1$ terms of the infinite series. To find the value of the infinite sum, he uses $(1)$:
$ \sum\limits_{k=0}^\infty {2^{k-1}\over (2^{k }+1)-(2^{k-1}+1)}=\lim_{n\rightarrow\infty} S_n =\lim_{n\rightarrow\infty} \Bigl[ {1\over 2^{0-1}+1}-{1\over 2^n -1}\Bigr]\cdots $