Question

What is the limit of (n!+2^n)^(1/n) as n->infinity?

Answer 1

just to check...\[\lim_{n \rightarrow \infty} (n!+2^n)^\frac{1}{n}\]Is that right?

Answer 2

Indeed. I do not understand why the answer is infinity. I have tried to use the Squeeze Theorem, and got a finite value.. :/

Answer 3

Huh...I can't think of anything. The inside argument (n!+2^n) would go to infinity very quickly

Answer 4

Yes, but is there any mathematical solution for this, and not just intuition? :)

Answer 5

Graphing it: http://awesomescreenshot.com/00a2fffj55

Answer 6

Even on a test? :D

(this question was taken from a test)

Answer 7

Hmmm...(2^x)^(1/x) turns into just 2 (except at x=0)
I don't really know

Answer 8

Because the factorial is the most powerful operation. So, you only need to pay attention to what it does as it goes to infinity

Answer 9

@donny471 what does it mean that factorial is most powerful?

Answer 10

Just as exponentials have more of an effect than polynomials, factorials dominate over both of them

Answer 11

So, we basically ignore the other operations since they offer a minimal influence on the overall behavior of the function

Answer 12

Exponentials are the fastest growing to infinity.. way faster than factorials..

Answer 13

Think of 2^10 vs 10!, even for a low n which is 10, the logarithem is much much bigger..

Answer 14

10! is drastically larger than 2^10

Answer 15

OMG, MY BAD -_-

Answer 16

If we arrange things, which is the fastest growing to infinity? (and the slowest).. I know that ln is really slow, but do you know any other important ones?

1. x^n (x>0)
2. n!
3. ln(n)
4. n^n

Answer 17

(i just listed a few, the order has no meaning)

Answer 18

the n^n is most powerful there

Answer 19

If you're ever uncertain just plug values into a calculator

Answer 20

lim of as n->infinity.
Why does the factorial lose here?

Answer 21

plug in 2 for n to start. Then 20... Whichever one overflows your calculator first is more powerful.

Answer 22

Really it's whichever one grows faster. If you have a dingy calculator it'll overflow quickly though

Answer 23

the problem is that we are not allowed to use calculators in test.. and this one is taken from a test :/

Answer 24

Use Sterling's formula to deduce that
\[
n!\approx \left(\frac{n}{e}\right)^n
\sqrt{2 n \pi
}>\left(\frac{n}{e}\right)^
n \\Hence\\
(n!)^{1/n}>\frac{n}{e}\\
\lim_{n\to \infty } (n!)^{1/n} > \lim_{n\to \infty }\frac{n}{e}=\infty\\
\]

Answer 25

Your limit is bigger than the above. So you are done.
See Stirling's formula on
http://en.wikipedia.org/wiki/Stirling%27s_approximation

Answer 26

Thanks eliassaab!

Answer 27

YW