<s>[INST] 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?
 [/INST] 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:
$$\sum_{k=0}^n\left(\frac{1}{2^{k-1}+1}-\frac{1}{2^k+1}\right)=\frac{1}{2^{0-1}+1}-\frac{1}{2^n+1}.$$
</s>