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!
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
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##?
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...
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##