a _{n}=((-1)^{n-1}(6)+2)/(2n+1+(-1)^{n-1})
find limit as n goes to infinity

Question

a _{n}=((-1)^{n-1}(6)+2)/(2n+1+(-1)^{n-1})
find limit as n goes to infinity

anonymous · Answer

$$a _{n}=((-1)^{n-1}(6)+2)/(2n+1+(-1)^{n-1})$$

inkyvoyd · Answer

THat's undefined

inkyvoyd · Answer

Infinity isn't necessarily even or odd

anonymous · Answer

$$\lim_{n\to \infty}\frac{(-1)^n6+2}{2n+1+(-1)^n}$$ like that ?

anonymous · Answer

no, satellite to the power of n-1

anonymous · Answer

$$\lim_{n\to \infty}\frac{(-1)^n6+2}{2n+1+(-1)^{n-1}}$$ like this?

anonymous · Answer

there is no "n" in the numerator, so as n goes to infinity so does the denominator. the numerator is either -4 or 8 but in any case the whole thing goes to zero

experimentx · Answer

http://www.wolframalpha.com/input/?i=limit+n-%3Einf+%28%28-1%29%5En+6+%2B+2+%29%2F%282n%2B1%28-1%29%5E%28n-1%29%29

anonymous · Answer

can I assume the limit  to infinity of the absolute value goes to zero? or does taking the absolute value break the alternating?