Using f(x) = lnx, and f(n-1) < Sn n-1 f(x) dx < f(n)

Show that for n>1

(n-1)! < n^n (n-1)! / (e(n-1)^(n-1)) < n!

Have tried to approach this using graphs, but could not get middle thing. Done this by adding the rectangle below f(x) and rectangles above f(x) with lengths of 1 unit square, starting with x =1

2. ## Clarification

To restate the problem:

Using $f(x)=\ln(x)$, and

$\displaystyle{f(n-1)<\int_{n-1}^{n}f(x)\,dx

show that for $n>1$, we have

$\displaystyle{(n-1)!<\frac{n^{n}(n-1)!}{e(n-1)^{n-1}}

Is that correct?

If so, I would compute the integral

$\displaystyle{\int_{n-1}^{n}f(x)\,dx=\int_{n-1}^{n}\ln(x)\,dx},$

and combine all the terms into one logarithm.

That should give you some direction.

3. Originally Posted by Lukybear
Using f(x) = lnx, and f(n-1) < Sn n-1 f(x) dx < f(n)

Show that for n>1

(n-1)! < n^n (n-1)! / (e(n-1)^(n-1)) < n!

Have tried to approach this using graphs, but could not get middle thing. Done this by adding the rectangle below f(x) and rectangles above f(x) with lengths of 1 unit square, starting with x =1
To start, I would take the logs of your three terms in the desired inequalities. Now what's $\ln[(n-1)!]$?

4. To both above posters thxs for replying. I have tried previously adding areas above the f(x) and below the f(x) but however, i do not get the middle term. Instead i get n^n/e^(n-1) when simplifying.

5. Are you sure you're simplifying correctly? For the integral I mentioned, I get

$\displaystyle{\ln\left(\frac{n^{n}}{e\,(n-1)^{n-1}}\right).}$

Can you show your line-by-line simplification of the integral?

6. Yea sorry my bad. Reading it wrongly, through the limits were n - 1

7. So have you gotten the needed result?