Monday, 30 March 2015

real analysis - Prove that the sequence zn, where zn:=xnyn, converges and lim(xnyn)=limzn=(limxn)(limyn)




Prove that the sequence zn, where zn:=xnyn, converges and lim



Also, when all limits when n\to\infty.



Here is what I have so far:
let x=\lim(x_n), y=\lim(y_n), z=x-y



let \epsilon>o, find an M_1 such that for all n\geq M_1, we have |x_n-x|<\frac{\epsilon}{2};find an M_2 such that for all n\geq M_2, we have |y_n-y|<\frac{\epsilon}{2}; take M:=\max\{M_1, M_2\}




For all n\geq M we have:



|z_n-z|=|(x_n-y_n)-(x-y)|=|x_n-x-y_n+y|≥ |x_n-x|-|y_n-y|



I don't know how to proceed from here, I want to get something like |z_n-z|<\epsilon, to complete the proof, right?



Also, is this sufficient enough to prove the second part of the question--the limit part



Please help me here. Thank you!


Answer




Hint:



You're very close! Note that:
|z_n-z|=|(x_n-y_n)-(x-y)|=|(x_n-x)-(y_n-y)|\leq|x_n-x|+|y_n-y|
It was the other side of the triangular inequality!



Please, use MathJax when typing math formulas! :)


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