Inequalities solved by considering areas

39 Views Asked by At

So basically, I've figured out the first two parts of the question, and I really need help with part b. I'm not sure how to tackle the question, nor where to begin. Can anyone help me? Please. All help is greatly appreciated :)

Sketch $y = \ln x, \: x \ge 1$
a. Using the areas of trapezia, show $\int_1^k \ln x \;dx > \frac12 \ln k +\ln \{(k-1)!\}$
b. Given $\int \ln x \;dx = x \ln x - x$, show that $k! < e\sqrt{k} \left( \dfrac{k}e \right)^k$

1

There are 1 best solutions below

8
On BEST ANSWER

Given $(b)$ and using in $(a)$, we have $$k \ln k - k+1 > \tfrac12 \ln k + \ln (k-1)!$$ $$\implies \ln \frac{k^k}{e^{k-1}\sqrt k} > \ln (k-1)! \implies (k-1)! < \frac{k^k}{e^{k-1} \sqrt k}$$ $$\implies k! < e\sqrt{k} \left(\frac{k}e \right)^k$$