geometric series

$\begin{align*} \sum\limits_{n=1}^\infty ar^{n-1} &= a+ar+ar^2+ar^3+\dots+ar^{n-1}+\dots \qquad a\neq 0 \\ &\boxed{=\frac{a}{1-r}=\frac{\text{first term}}{1-\text{common ratio}}} \qquad \text{ if } |r|<1 \end{align*}$

GS 2

By telescoping products
Assuming $0<r<1$ :

$\begin{align*} S&=1+r+r^2+r^3+r^4+r^5+\cdots \\ &=1(1+r)+r^2(1+r)+r^4(1+r)+\cdots \\ &=(1+r)(1+r^2+r^4+r^6+\cdots \\ &=(1+r)(1+r^2)(1+r^4+r^8+\cdots \\ &=(1+r)(1+r^2)(1+r^4)(1+r^8)(1+\cdots \\ S(1-r)&=(1-r^2)(1+r^2)(1+r^4)(1+r^8)(1+\cdots \\ &=(1-r^4)(1+r^4)(1+r^8)(1+r^{16})(1+\cdots \\ &=(1-r^8)(1+r^8)(1+r^{16})(1+\cdots \\ &\to 1 \\ S&=\frac{1}{1-r} \end{align*}$

Sum of the first $n$ terms
1.

$\begin{align*} S_n:1&=r^n-1:r-1 \\ S_n&=\frac{r^n-1}{r-1} \end{align*}$

$1-x^{n+1}=(1-x)(1+x+x^2+x^3+\dots+x^n).$

$\frac{1-x^{n+1}}{1-x}=1+x+x^2+x^3+\dots+x^n.$

and $(1-x^{n+1})/(1-x)$ is the generating function for the sequence $\angb{1,1,1\dots 1,0,0,0\dots}$ , where the first $n+1$ terms are $1$ .

$s_n=a+ar+ar^2+\dots+ar^{n-1}$

Multiply both sides by $r$ :

$rs_n=ar+ar^2+\dots+ar^{n-1}+ar^n$

Subtract the second equation from the first:

$s_n-rs_n=a-ar^n$

$\boxed{s_n=\frac{a(1-r^n)}{1-r}}$

Problem: Let $p$ , $q$ be two distinct primes. Prove that there are positive integers $a$ , $b$ so that the arithmetic mean of all the divisors of $p^a q^b$ is also an integer.
Solution: The sum of all divisors of $n=p^a q^b$ is given by

$(1+p+p^2+p^3+\cdots+p^a)(1+q+q^2+q^3+\cdots+q^b)$