Recent content by Seth|MMORSE

NONE of my previous exercises required me to use such a function ... never occurred to me ... till something struck. Anyway, thank you so much for your help!

##M=max(N,N_1)##? I kept thinking of using inequalities ...

I don't see anything in post #25 :confused: 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

##M=\frac{a_1+a_2+...+a_M}{\epsilon+a}##? From equating ##\epsilon## and ##b_M-a##

Therefore it should be there exists an ##M\geq N## such that ##|b_n-a|<\epsilon\ \forall n>M##?

Since for ##n>N##, we are able to get ##|b_n-a|<\epsilon## ... Therefore, if ##n>M\geq N, |b_n-a|<\epsilon##

##M>N_1## that you defined? I have the feeling that I have to start learning analysis from scratch again ...

Like the one stated in the workings? ##n>## than something for the first section and just ##>N## for the second one. I can visualise ##a_n## eventually lying within ##\epsilon/2##'s after ##N##, but how would I know where M would be for ##b_n##?

Does it have anything to do with the fact that I used ##N## for the first sequence and I used ##\frac{\epsilon}{2}## as a bound?

Suppose ##(a_n)\rightarrow a##. Show that ##(b_n)\rightarrow a## when ##b_n=\frac{a_1+a_2+...+a_n}{n}## Definition of ##(a_n)\rightarrow a##: For each ##\epsilon>0, \exists N\in \mathbb{N}:|a_n-a|<\frac{\epsilon}{2} \forall n>N## Show ##\exists M\in \mathbb{N}:|b_n-a|<\epsilon\ \forall...

Suppose ##(a_n)\rightarrow a##. Show that ##(b_n)\rightarrow a## when ##b_n=\frac{a_1+a_2+...+a_n}{n}## Definition of ##(a_n)\rightarrow a##: For each ##\epsilon>0, \exists n\in \mathbb{N}:|a_n-a|<\epsilon\ \forall n>N## Let ##|a_n-a|<\frac{\epsilon}{2}\ \forall n>N_0## Show...

Oh, they're just ##\epsilon##'s that correspond to each term ... which are essentially the usual ##\epsilon##'s. Sorry about that.

##\frac{1}{n}(|a_1-a|+...+|a_N-a|)<\frac{\epsilon}{2}## this? ##\frac{1}{n}(|a_1-a|+...+|a_N-a|)<\frac{\epsilon}{2}## ##\frac{2(|a_1-a|+...+|a_N-a|)}{\epsilon}<n##

Okay, thought about it again (okay not really, kinda discussed with my friends) and this is the result. Let ##\epsilon>0, |a_n-a|<\frac{\epsilon}{2} \ \forall n>N##...

Only thing I could come up with is this: ##(|\frac{(a_1-a)}{n}|+...+|\frac{(a_N-a)}{n}|)<\frac{|a_n-a|}{2}<\frac{\epsilon}{2}## ##\frac{N(a_N-a)}{n}<\frac{|a_n-a|}{2}## ##\frac{N}{n}<\frac{1}{2}## ##n>2N##