Explanation in a step of the summation $\frac{1}{2^n+1} + \frac{1}{2^n+2}+\cdots +\frac{1}{2^{n+1}}$

Question

I have the solution to the above equation but I wanted to know the explanation for it.

$$\frac{1}{2^n+1} + \frac{1}{2^n+2}+\cdots +\frac{1}{2^{n+1}} = 2^n \frac{1}{ 2^{n+1}}$$ and $$2^n\frac{1}{2^{n+1}}=\frac{1}{2} \qquad \text{(I understand this part)}.$$

I wasn't sure how that became the product of those 2 numbers. Any explanation please? Thank you.

To make it clearer let me include this link: https://www2.isye.gatech.edu/~hsharp/math2420/harmonic.pdf

Look at line 5

Im open to any alternative suggestions to solve the problem.

$\frac{1}{2^{n+1}}$ is the smalest term of your sum, and you have $2^n$ terms, so your sum is bounded by number of term multiplied by smalest term, which is what you wrote. Conversely you can show that it is bounded by above by number of term times largest one which turns up to be $1$. so your sum is between $1/2$ and $1$. — P. Quinton, Oct 09 '20 at 09:33
I hazard a guess that you really have this sum with $k$ there equal to your $2^n$. Anyway, there is no closed form for this sum, but as $n\to\infty$ it tends to $\ln 2$ as a limit. In other words, the "equality" of your first displayed formula is not correct. — Jyrki Lahtonen, Oct 09 '20 at 09:35
Try $n=1$ or $n=2$ for example, the equality is not correct. — Fareed Abi Farraj, Oct 09 '20 at 09:36
The pdf is not wrong as it has a $\ge$ sign and not an $=$ at the start of line 5. You incorrectly changed it to an equals sign here. — Jaap Scherphuis, Oct 09 '20 at 11:08

score 1 · Answer 1 · answered Oct 09 '20 at 09:40

1

We have an inequality since $$\frac{1}{2^n+i}\geq \frac{1}{2^{n+1}}$$

for $1\leq i\leq 2^{n}.$ Thus $$\sum_{i=1}^{2^{n}}\frac{1}{2^n+i}\geq\sum_{i=1}^{2^n}\frac{1}{2^{n+1}}=\frac{1}{2^{n+1}}\sum_{i=1}^{2^n}1=2^n\frac{1}{2^{n+1}}$$

answered Oct 09 '20 at 09:40

Alessio K

10,599

Explanation in a step of the summation $\frac{1}{2^n+1} + \frac{1}{2^n+2}+\cdots +\frac{1}{2^{n+1}}$

1 Answers1