# Show that $\frac{1}{2} +\frac{1}{2^2} +\frac{1}{2^3} + … +\frac{1}{2^n} < 1$ [duplicate]

Show that $\frac{1}{2} +\frac{1}{2^2} +\frac{1}{2^3} + ... +\frac{1}{2^n} < 1$ and explain how the picture below is a good representation of this inequality.

I came across this in an Algebra book and I'm really interested in how the proof for this problem looks like. Does anyone know how to go about it ?

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• The area of the square is $1$. Now find the areas of the rectangles inside the square and add them. – Math Lover Dec 7 '17 at 22:48

The inequality is easy to show since it's simply a geometric series.

For the image, assume that each side of the square has length one. Note that the left-most rectangle is 1/2 of the total area, the bottom right square is 1/4 of the total area, and so on. Each smaller rectangle is half the size of the previous one. That makes it a pictorial representation of the LHS of the inequality.

Hint:

$$u+u^2+\dots+u^n=u(1+u+\dots+u^{n-1})=u\,\frac{1-u^n}{1-u}=\frac u{1-u}\color{red}-\frac{u^{n+1}}{1-u}.$$

My answer is an elaboration on user3727079's. So, put the upvote there. Imagine the whole picture has a value of $1$. Then, the biggest rectangle has a value of $\frac{1}{2}$ because it is one half of the whole. Thus you get $$\frac{1}{2} < 1,$$ because the biggest square does not fill up the picture. Now, the next biggest rectangle is only one fourth of the whole picture. So, it only has a value of $\frac{1}{4}$, which is $\frac{1}{2^2}$. We put this rectangle together with the previous, biggest rectangle, and we still don't get the whole. So $$\frac{1}{2} + \frac{1}{2^2} < 1.$$ Then the next rectangle would have a value of $\frac{1}{8}$ because it is one eighth of the whole picture. And, $\frac{1}{8}$ is $\frac{1}{2^3}$. But, adding on this rectangle doesn't make the whole picture. Thus $$\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} < 1.$$ No matter how many rectangles we add on, we never get the whole picture. That is what is meant by $$\frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^n} < 1.$$

$\frac 12 + \frac 1{2^2} + .... + \frac 1{2^{n-1}} + \frac 1{2^n} = k$

$\frac 12 + \frac 1{2^2} + .... + \frac 1{2^{n-1}} + \frac 1{2^n} + \frac 1{2^n}= k + \frac 1{2^n}$

$\frac 12 + \frac 1{2^2} + .... + \frac 1{2^{n-1}} + \frac 2{2^n}= k + \frac 1{2^n}$

$\frac 12 + \frac 1{2^2} + .... + \frac 1{2^{n-1}} + \frac 1{2^{n-1}}= k + \frac 1{2^n}$

$\frac 12 + \frac 1{2^2} + .... + \frac 2{2^{n-1}} = k + \frac 1{2^n}$

$\frac 12 + \frac 1{2^2} + .... + \frac 1{2^{n-2}} + \frac 1{2^{n-2}}= k + \frac 1{2^n}$

$\frac 12 + \frac 1{2^2} + .... + \frac 2{2^{n-2}} = k + \frac 1{2^n}$

$.....$

$\frac 12 + \frac 1{2^2} + \frac 1{2^3} + \frac 1{2^3} = k + \frac 1{2^n}$

$\frac 12 + \frac 1{2^2} + \frac 2{2^3} = k + \frac 1{2^n}$

$\frac 12 + \frac 1{2^2} + \frac 1{2^2} = k + \frac 1{2^n}$

$\frac 12 + \frac 2{2^2} = k + \frac 1{2^n}$

$\frac 12 + \frac 12 = k + \frac 1{2^n}$

$1 = k + \frac 1{2^n}$

$k = 1 - \frac 1{2^n}$.

.... or ....

$\frac 12 + \frac 1{2^2} + .... + \frac 1{2^{n-1}} + \frac 1{2^n} = k$

$2^n(\frac 12 + \frac 1{2^2} + .... + \frac 1{2^{n-1}} + \frac 1{2^n}) = 2^n*k$

$2^{n-1} + 2^{n-2} + ...... + 2^2 + 2 + 1 = 2^n*k$

$2^{n-1} + 2^{n-2} + ...... + 2^2 + 2 + 1 + 1 =2^n*k+1$

$2^{n-1} + 2^{n-2} + ...... + 2^2 + 2 + 2 =2^n*k+1$

$2^{n-1} + 2^{n-2} + ...... + 2^2 + 4 =2^n*k+1$

$2^{n-1} + 2^{n-2} + ...... + 8 =2^n*k+1$

$....$

$2^{n-1} + 2^{n-2} + 2^{n-2} = 2^n*k + 1$

$2^{n-1} + 2^{n-1} = 2^n*k + 1$

$2^n = 2^n*k + 1 > 2^n*k$

$1 = \frac {2^n}{2^n} = \frac {2^n}{2^n}*k + \frac 1{2^n} = k + \frac 1{2^n} > k$.

..... or ......

$k = \sum_{i=1}^n \frac 1{2^i}$

$2k = 2\sum_{i=1}^n \frac 1{2^i} = \sum_{i=1}^n \frac 2{2^{i}}=\sum_{i=1}^{n}\frac 1{2^{i-1}} = \sum_{i=0}^{n-1}\frac 1{2^i}$.

$2k = \sum_{i=0}^{n-1} \frac 1{2^i} = \frac 1{2^0} + \sum_{i=1}^n\frac 1{2^i} - \frac 1{2^n} = 1 + k - \frac 1{2^n}$

$k = 1-\frac 1{2^n} < 1$.

To prove the inequality, we can show a stronger result.

Define $s(n):= \frac 12 + \frac {1}{2^2} + \frac {1}{2^3} + \cdots + \frac {1}{2^n}$ . Then $1-s(n) = \frac {1}{2^n}$.

Proof by induction:
Base case: $1-s(1) = 1-\frac 12 = \frac 12 = \frac 1{2^1}$ as required.

Assume $\color{red}{1-s(k) = \frac {1}{2^k}}$ as the inductive hypothesis.

Then $s(k+1) = s(k) + \frac 1{2^{k+1}}$ from the definition,
and $1-s(k+1) = \color{red}{1-s(k)}-\frac 1{2^{k+1}} = \color{red}{\frac 1{2^k}}-\frac 1{2^{k+1}} = \frac 1{2^{k+1}}$ as required, completing the proof.

Back to the inequality: $1-s(n)>0$ and thus $s(n)<1$.

As noted by others, the diagram shows a progression of rectangles with the area halving for each next smaller rectangle, which for the outer square of area $1$ would demonstrate the relationships proven above.