Proving the Limit of a Sequence Math Problem with Continuous Function f(x)

[HACR]

Homework Statement

Prove that if f(x) is continuous for 0<f(x)<1, then $$lim_{n->\infty}\frac{1}{n}[(n+1)(n+2)(n+3)...(2n)]^{\frac{1}{n}}=\frac{4}{e}$$.

Homework Equations

f(x)=log(1+x)

The Attempt at a Solution

We know that $$lim_{n->\infty}\frac{1}{n}[f(\frac{1}{n})+f(\frac{2}{n})+f(\frac{3}{n})+...f(\frac{n}{n})]=\int_0^1 f(x)dx$$

So that $$\int_0^1 log(1+x)(=f(x))dx$$ equals the sequence in the limit.
But evaluation of the integral shows that it is a divergent one.

 

[Char. Limit]

There's no way that integral can be divergent. In the interval between 0 and 1, log(1+x) is bounded by 0 and log(2), respectively. So the integral of log(1+x) needs to be between 0 and log(2). It should be relatively easy to prove that the integral is bounded above and below by those two numbers.

 

[HACR]

Right, I should've fixed the limits of integration from 0to 1 to 1 to 2 when i did it. Thanks for point it out. Although I think the 1/n power indicated in the problem is not necessary.

 

[Char. Limit]

So after you've found the value of the integral, what's your next step? Do you know?

 

[HACR]

It turned out to be $$\lim_{n->\infty}\frac{1}{n}(n+1)(n+2)(n+3)...(2n)=\frac{e}{4}$$

 

[Char. Limit]

Hmm, but did you remember the 1/n power? First thing I'd try doing is taking the logarithm of the left and right sides.

 

[HACR]

oops.