# Sequence math problem

1. Jan 12, 2012

### HACR

1. The problem statement, all variables and given/known data

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

2. Relevant equations

f(x)=log(1+x)
3. 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.

2. Jan 12, 2012

### Char. Limit

Re: sequence

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.

3. Jan 13, 2012

### HACR

Re: sequence

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.

Last edited: Jan 13, 2012
4. Jan 13, 2012

### Char. Limit

Re: sequence

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

5. Jan 13, 2012

### HACR

Re: sequence

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

6. Jan 13, 2012

### Char. Limit

Re: sequence

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.

7. Jan 13, 2012

Re: sequence

oops.