- #36
Seth|MMORSE
- 22
- 0
##M=\frac{a_1+a_2+...+a_M}{\epsilon+a}##?
From equating ##\epsilon## and ##b_M-a##
From equating ##\epsilon## and ##b_M-a##
LCKurtz said:No. The best I can do without giving you the answer, which I'm not going to do, is tell you to re-read post #25. You are stuck on a really easy issue, unfortunately.
Seth|MMORSE said:I don't see anything in post #25
I'm can't help it that my brain doesn't know what to look for, just staring at it over and over again without seeing anything different ... sigh
Seth|MMORSE said:##|b_n-a|##
##= |\frac{a_1+a_2+...+a_n}{n}-a|##
##=|\frac{a_1+a_2+...+a_n-na}{n}|##
##=|\frac{(a_1-a)+(a_2-a)+...+(a_n-1)}{n}|##
By triangle inequality,
##\leq |\frac{a_1-a}{n}|+|\frac{a_2-a}{n}|+...+|\frac{a_n-a}{n}|##
##=\underbrace{(|\frac{a_1-a}{n}|+...+|\frac{a_{N}-a}{n}|)}_{p}+
\underbrace{(|\frac{a_{N+1}-a}{n}|+...+|\frac{a_n-a}{n}|)}_{q}##
##p=|\frac{a_1-a}{n}|+...+|\frac{a_{N}-a}{n}|##
##=\frac{1}{n}(|a_1-a|+...+|a_{N}-a|)##
We define ##n>\frac{2(|a_1-a|+...+|a_{N}-a|)}{\epsilon}##
##\frac{1}{n}(|a_1-a|+...+|a_{N}-a|)<\frac{\epsilon}{2}##
##\Rightarrow p<\frac{\epsilon}{2}##
##q=|\frac{a_{N+1}-a}{n}|+...+|\frac{a_n-a}{n}|##
Using ##|a_n-a|<\frac{\epsilon}{2}\ \forall n>N##
##q<\frac{\epsilon/2}{n}+...+\frac{\epsilon/2}{n}##
##=(\frac{n-N}{n})(\frac{\epsilon}{2})##
##=(1-\frac{N}{n})(\frac{\epsilon}{2})##
Since ##n>N, (1-\frac{N}{n})<1##
##(1-\frac{N}{n})(\frac{\epsilon}{2})<\frac{\epsilon}{2}##
##\Rightarrow q<\frac{\epsilon}{2}##
##|b_n-a|=p+q##
##|b_n-a|< \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon##
##|b_n-a|<\epsilon##
Seth|MMORSE said:##M=max(N,N_1)##?
I kept thinking of using inequalities ...
LCKurtz said:There now. That wasn't so hard, was it? I hope you have learned a few things about ##\epsilon,N## proofs in this thread. So you will just breeze through your next such problem.