Thread: limit of (2^n)/(n(n+1)(n+2))

1. limit of (2^n)/(n(n+1)(n+2))

How would I know that the limit of (2^n)/(n(n+1)(n+2)) as n tends to infinity is 0?

2. Re: limit of (2^n)/(n(n+1)(n+2))

It isn't. It diverges to infinity.

3. Re: limit of (2^n)/(n(n+1)(n+2))

WolframAlpha says it is 0.

(2^n)/(n(n+1)(n+2)) - Wolfram|Alpha

4. Re: limit of (2^n)/(n(n+1)(n+2))

Originally Posted by Stuck Man
WolframAlpha says it is 0.

(2^n)/(n(n+1)(n+2)) - Wolfram|Alpha
No it does not

5. Re: limit of (2^n)/(n(n+1)(n+2))

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

6. Re: limit of (2^n)/(n(n+1)(n+2))

Wolframalpha says the limit as n tends to NEGATIVE infinity is zero (which is true). The numerator becomes arbitrarily close to zero while the absolute value of the denominator grows arbitrarily large.

7. Re: limit of (2^n)/(n(n+1)(n+2))

Originally Posted by Stuck Man
WolframAlpha says it is 0.

(2^n)/(n(n+1)(n+2)) - Wolfram|Alpha
The first graph given is misleading, the second graph (which is zoomed out) gives a better picture...

8. Re: limit of (2^n)/(n(n+1)(n+2))

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.

9. Re: limit of (2^n)/(n(n+1)(n+2))

Originally Posted by Stuck Man
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.
Let p be any polynomial such that the Limit as n approaches infinity of p(n) is infinity. Then the Limit as n approaches infinity of a^n/p(n) is infinity for any a>1.