Basic Summation

$\begin{aligned} S & = \sum_{j=1}^\infty 2^{-j} \\ & = \frac 12 + \frac 14 + \frac 18 + \cdots \\ \implies 2S & = 1 + \blue{\frac 12 + \frac 14 + \frac 18 + \cdots} \\ & = 1 + \blue S \\ \implies S & = \boxed 1 \end{aligned}$

$\begin{aligned}\sum_{j=1}^\infty 2^{-j}&=2^{-1}+2^{-2}+2^{-3}+...\\&=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...\end{aligned}$

This is an infinite geometric series with first term $\dfrac{1}{2}$ and common ratio $\dfrac{1}{2}$ . Since $\left |\dfrac{1}{2}\right |< 1$ the sum converges, and it converges to $\dfrac{\frac{1}{2}}{1-\frac{1}{2}}=\dfrac{0.5}{0.5}=\color{#20A900}{\boxed{1}}$ .

Proof that for a geometric series with first term $a$ and common ratio $r$ , the sum to infinity is $\dfrac{a}{1-r}$ :

Let $S_n$ be the sum of the first $n$ terms.

$\begin{aligned}S_n&=a+ar+ar^2+ar^3+...+ar^{n-1}\\\implies rS_n&=ar+ar^2+...+ar^n\\\iff rS_n-S_n&=ar^n-a\\\iff S_n(r-1)&=a(r^n-1)\\\iff S_n&=\frac{a(1-r^n)}{1-r}\end{aligned}$

Let $n\to\infty$ . If $|r|<1$ then $r^n\to 0$ , so the sum to infinity $S_\infty=\lim_{n\to\infty} S_n=\dfrac{a\cdot (1-0)}{1-r}=\dfrac{a}{1-r}$ .

$\displaystyle\sum_{j=1}^{\infty}2^{-j}=\left ( \displaystyle\sum_{j=0}^{\infty} 2^{-j}\right ) -1=\dfrac{1}{1-\dfrac{1}{2}}-1=2-1=\boxed{1}$

In general, for any $\displaystyle\sum_{j=0}^{\infty}a^j$ , with -1<a<1,

$\displaystyle\sum_{j=0}^{\infty}a^j=\dfrac{1}{1-a}$

Proof:

$\begin{aligned} S &= 1+a+a^2+... \\S-1&=a+a^2+...\\ \dfrac{S-1}{a}&=1+a+a^2+... =S\\ \dfrac{S-1}{a}&=S\\ S-1&=aS\\ S(1-a)&=1\\S&=\dfrac{1}{1-a}\ \blacksquare\end{aligned}$

| $a$ | must be less than one because, if it weren't like that, the sum wouldn't converge.

Then, we can generalize, and use the property of

$\displaystyle\sum_{j=x}^n a^j = \dfrac{a^x-a^{n+1}}{1-a}$

Which I'm proving now:

$\begin{aligned} S &= a^x+a^{x+1}+...+a^n \\ aS &= a^{x+1}+a^{x+2}+...+a^{n+1} \\ aS-S &= a^{n+1}-a^x \\ S(a-1) &= a^{n+1}-a^x \\ S &= \dfrac{a^{n+1}-a^x}{a-1} = \dfrac{a^x-a^{n+1}}{1-a} \ \blacksquare \end{aligned}$

Here, $a$ doesn't have restrictions, because as it's a finite sum, it always converges. However, when taking the limit, we have the same restriction.

Now, we take the limit:

$\begin{aligned}\lim_{n\to\infty} \displaystyle\sum_{j=1}^n 2^{-j} &= \lim_{n\to\infty} \dfrac{2^{-1}-2^{-(n+1)}}{1-2^{-1}} \\ &=\dfrac{2^{-1}}{1-2^{-1}} \\ &= 2^{-1}\times 2 \\ &= \boxed{1} \ \ \blacksquare \end{aligned}$

This is equal to: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...$ This is an infinite geometric series with common ratio $\frac{1}{2}$

Using the sum of infinite geometric series formula, our answer is: $\frac{\frac{1}{2}}{\frac{1}{2}} = \boxed{1}$

$S=\displaystyle \sum_{j=1}^\infty 2^{-j} =2^{-1}+2^{-2}+2^{-3}+\ldots =\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\ldots$

$\text{Some mathematical law thingy states that: } \frac{1}{x^1}+\frac{1}{x^2}+\frac{1}{x^3}+\ldots=\frac{1}{x+1}$

$\text{In this case, } x=2$

$S=\frac{1}{x^1}+\frac{1}{x^2}+\frac{1}{x^3}+\ldots = \frac{1}{x+1} \Rightarrow \frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\ldots=\frac{1}{1}$

$S=\frac{1}{1}=\boxed{1}$