Evaluating $\sum_{n=0}^{\infty} \left(\frac{1}{2^{n-1}+1 }+ \frac{1}{2^n+1}\right)$ Evaluating $$\sum_{n=0}^{\infty} \left(\frac{1}{2^{n-1}+1 }+ \frac{1}{2^n+1}\right)$$
Full Question

Provided Answer 


But how do I get from 
$$\sum \frac{1}{2^{k-1}+1} - \frac{1}{2^k+1} = \frac{1}{2^{\color{red}0-1}+1}...$$
Why is the summation removed. And why is $k=0$? 
In the next line it becomes $n$ again? Then $\lim$ was introduced? Why?
 A: In the "provided answer," the first two sums should go to $n$, not to infinity. To see how the summation was removed, write out the first few terms of the summation, and note the cancellation that takes place. 
A: It is a finite telescoping sum. For any sequence $\rm\{ x_k\}_{k=1}^\infty$ we have
$$\rm \begin{array}{c l} \sum_{k=a}^b (x_k-x_{k+1}) & \rm =(x_a-\color{Maroon}{x_{a+1}})+(\color{Maroon}{x_{a+1}}-x_{a+2})+\cdots+(x_{b-1}-\color{Purple}{x_b})+(\color{Purple}{x_b}-x_{b+1} ) \\ & \rm =x_a-x_{b+1}. \end{array}$$
Notice the repeated cancellation? We could also shift the index back by one, as here:
$$\sum_{k=0}^n\left(\frac{1}{2^{k-1}+1}-\frac{1}{2^k+1}\right)=\frac{1}{2^{0-1}+1}-\frac{1}{2^n+1}.$$
A: 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 $(1)$:
$$
 \sum\limits_{k=0}^\infty {2^{k-1}\over (2^{k }+1)-(2^{k-1}+1)}=\lim_{n\rightarrow\infty} S_n =\lim_{n\rightarrow\infty} \Bigl[ {1\over 2^{0-1}+1}-{1\over 2^n -1}\Bigr]\cdots
$$
