How do I find the partial sum of this

Question

So I have a sum defined below:

$$ \sum_{m=1}^n 2^{-m} $$

I know the partial sum equals

$$ \frac{1}{2^n}(2^n - 1)\ $$

But how do you go from one to the other?

Answer 1

Let $$S=\sum_{k=1}^n 2^{-k}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots +\frac{1}{2^n}.\tag{1}$$ Then $$2S=1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^{n-1}}.\tag{2}$$ From (1) and (2), by subtraction we obtain $$S=1-\frac{1}{2^n}$$ (note the massive amount of cancellation). Now we are almost at the desired expression.

Answer 2

By Taylor series you can find $\forall x : |x|<1$

$$\frac{1}{1-x}=\sum_{k=0}^{\infty} x^k$$

And so

$$\sum_{k=0}^{m} x^k=\sum_{k=0}^{\infty} x^k - \sum_{k=m+1}^{\infty} x^k=(1-x^{m+1})\sum_{k=0}^{\infty} x^k $$

$$=\frac{1-x^{m+1}}{1-x}$$

Answer 3

Hint: $$\sum_{k=0}^{n}x^k=\frac{1-x^{n+1}}{1-x}\ ,$$ hence $$\sum_{k=1}^{n}x^k=\frac{1-x^{n+1}}{1-x}-1\ .$$

You can obtain the first formula as following $$(1-x)\sum_{k=0}^{n}x^k=1-x+x-x^2+x^2-x^3+x^3+\cdots-x^n+x^n-x^{n+1}=1-x^{n+1}$$

Answer 4

A geometric sequence is a sequence $x_n$ where successive terms are obtained by multiplying the previous term by a fixed number $r$ - i.e. $$x_{n+1} = rx_n$$

For example, in your case we have $$\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16},\ldots$$ where we multiply by $\frac{1}{2}$ to get to the next term.

Given a sequence like this, it is possible to derive a simple formula to calculate the sum:

Let's say I want to calculate $$S_n = x_1 + x_2 + ... + x_n = x_1 + rx_1 + ... + r^{n-1}x_1$$

Then $$rS_n = rx_1 + r^2x+1 + \ldots r^{n}x_1$$ And subtracting the two gives $$S_n(1 - r) =x_1 - r^{n}x_1$$

From here we derive the formula $$S_n = \frac{x_1(1 - r^{n})}{1-r}$$

In your case, $x_1 = \frac{1}{2}$, $r = \frac{1}{2}$ and we want $$S_n = \frac{\frac{1}{2}(1-\frac{1}{2^n})}{1-\frac{1}{2}} = 1 - \frac{1}{2^n}$$

Answer 5

Given

$$ S_{[n]} = \sum_{m=1}^n 2^{-m}. $$

Then we can write

$$ 2 \color{green}{S_{[n]}} = 2 \underbrace{ \color{blue}{\sum_{m=1}^n 2^{-m}}}_{\displaystyle \color{green}{S_{[n]}}} = \sum_{m=1}^n 2^1 \cdot 2^{-m} = \sum_{m=1}^n 2^{1-m} = \sum_{m=0}^{n-1} 2^{-m} = \color{red}{\underbrace{1}_{m=0}} + \underbrace{\color{blue}{\sum_{m=1}^n 2^{-m}}}_{\displaystyle \color{green}{S_{[n]}}} \color{red}{- \underbrace{2^{-n}}_{m=n}}, $$

so we obtain

$$ 2 S_{[n]} = S_{[n]} + 1 - 2^{-n}, $$

so

$$ S_{[n]} = 1 - 2^{-n}. $$

Note that

$$ 1 - 2^{-n} = 2^{-n} \Big( 2^n - 1 \Big) = \frac{1}{2^n} \Big( 2^n - 1 \Big) $$