Question

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

Answer 1

\[a _{n}=((-1)^{n-1}(6)+2)/(2n+1+(-1)^{n-1})\]

Answer 2

THat's undefined

Answer 3

Infinity isn't necessarily even or odd

Answer 4

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

Answer 5

no, satellite to the power of n-1

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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