How would I know that the limit of (2^n)/(n(n+1)(n+2)) as n tends to infinity is 0?
WolframAlpha says it is 0.
(2^n)/(n(n+1)(n+2)) - Wolfram|Alpha
Hello, Stuck Man!
How would I know that the limit of (2^n)/(n(n+1)(n+2)) as n tends to infinity is 0?
I don't understand how Wolfram came up with that . . . but it's wrong!
An exponential function will outgrow a cubic function.
Darn! . . . LaTex isn't working!!
. . . . . . . . . . . . . . .2048
At n = 11, we have: ------
. . . . . . . . . . . . . . .1716
. . . . . . . . . . . . . . .1,048,576
At n = 20, we have: -----------
. . . . . . . . . . . . . . . . 9,240
I misread WolframAlpha. It gives the limit as n tends to negative infinity. If its true that in general an exponential function outgrows a cubic function or one with a higher degree then I will know what the limit is in future. That's assuming the base is greater than 1.
The graphs omit the growth pattern that starts getting noticeable at about x=8.