Question

sums ,products and limits

Answer 1

Given \(0<a<1\) and \(\large \sigma(m)=\sum_{k=0}^ma^k\)
show that

\(\color{blue}{\large \lim_{n\to \infty}\rho(n)=0}\)

if furthermore:

\(\large \rho(n)=\prod_{m=0}^n (1-\sigma(m)+a\sigma(m))\)

Answer 2

isn't that geometric series?

Answer 3

yes it is,i made this one up,so its just a challenge :)

Answer 4

\[ 1-\frac{ 1-a^{m+1} }{1-a} + a\frac{ 1-a^{m+1} }{1-a} = 1 + a^m\]

Answer 5

\[=a^m\] if you factorise

Answer 6

ok ok .. algebra mistake.

Answer 7

the limit is obvious!! just take b = 1/a

Answer 8

\[1-\sigma(m)+a\sigma(m)=1-(1-a)\sigma(m)\]

Answer 9

okay okay