Sunday, October 16, 2016

real analysis - Show that $(x_n) $ convergent to 0



Show that if $x_n \geq 0$ and the limit of $((-1)^nx_n) $ exists, then $(x_n) $ convergent to 0.




I don't have a single clue to solve the problem. I have looking back about monotone sequence, Cauchy sequence, and bounded one, but seem don't lead to this. Please help. Regards


Answer



Since $a_n\ge 0$ we have, $a_n = |(-1)^n a_n|$. If a sequence converges to L then its absolute value convergers to |L|, due to the continuity of the absolute value function.



In your case $L=0$ due to the oscillation of $ (-1)^n a_n.$


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