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.
geom series

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

2.

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

So

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

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.

3.

    \[s_n=a+ar+ar^2+\dots+ar^{n-1} \]

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

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

as can be seen by expanding the brackets. The number n has (a+1)(b+1) positive divisors, and their arithmetic mean is

    \[M=\frac{(1+p+p^2+p^3+\cdots+p^a)(1+q+q^2+q^3+\cdots+q^b)}{(a+1)(b+1)}.\]

If p and q are both odd, then when a=p and b=q,

    \[M=(1+p^2+p^4+p^6+\cdots+p^{p-1})(1+q^2+q^4+q^6+\cdots+q^{q-1}).\]

If p=2, choose b=q and a=q^2+q^4+q^6+\cdots+q^{q-1}.
Then M=1+2+2^2+\cdots+2^a}. (And similarly if q=2).


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