Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Full Question

enter image description here

Provided Answer

enter image description here


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?

share|improve this question
1  
I think you are missing a minus sign between the two terms. –  copper.hat Apr 20 '12 at 3:54
add comment

3 Answers

up vote 4 down vote accepted

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.

share|improve this answer
    
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
1  
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
add comment

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:

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

share|improve this answer
add comment

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 $(1)$:

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.