Monday, 12 May 2014

real analysis - limntoinftyfrac1nsumnk=1xk=x; given ;limntoinftyxn=x;?



I have a question which is giving me a hard time.



I want to show that limn1nnk=1xk=x given that limnxn=x.


Answer



Or you could slog through a tedious proof:



Choose ϵ>0. Let N be such that nN means |xnx|<ϵ2. Now choose NN so that nN means 1nNk=1|xnx|<ϵ2.




Then, if nN, we have the estimate:



|1nnk=1(xnx)|1nnk=1|xnx|1nNk=1|xnx|+1nnk=N+1|xnx|<ϵ2+n1nϵ2=ϵ



Hence limn1nnk=1(xnx)=0 from which the result follows.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...