# 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

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?

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I think you are missing a minus sign between the two terms. –  copper.hat Apr 20 '12 at 3:54

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.

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Another thing for part iii is why not just factorize out the $2$ in the 1st step? $2 \sum \frac{2^{n-1}}{...}$? But I will get a different answer tho: $2\cdot\frac{2}{3}=\frac{4}{3}$ –  Jiew Meng Apr 20 '12 at 4:42
Notice that in (iii) the summation starts at $n=1$, so it's $2\cdot{2\over3}$ minus the $(n=0)$-term. –  Gerry Myerson Apr 20 '12 at 4:55

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:

$$\sum_{k=0}^n\left(\frac{1}{2^{k-1}+1}-\frac{1}{2^k+1}\right)=\frac{1}{2^{0-1}+1}-\frac{1}{2^n+1}.$$

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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 $(1)$:

$$\sum\limits_{k=0}^\infty {2^{k-1}\over (2^{k }+1)-(2^{k-1}+1)}=\lim_{n\rightarrow\infty} S_n =\lim_{n\rightarrow\infty} \Bigl[ {1\over 2^{0-1}+1}-{1\over 2^n -1}\Bigr]\cdots$$

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