Estimate the factorial $n!$ starting with the integral of $1/x$

Question

This is a 3-part problem concerning an estimate for the factorial $n!$

a. By considering the graph of $y=\frac{1}{x}$ explain why $$\frac{1}{k+1} < \int\limits_{k}^{k+1} \frac{\mathrm dx}{x} < \frac{1}{k}$$ where $ k \in \mathbb Z$ is a positive integer

I figured (a) by considering the graph, but still added it for completeness to answer (b) and (c).

b. Deduce that $(1+\frac{1}{k})^k < e < (1+\frac{1}{k})^{(k+1)}$ for $k=1,2,3,..$

Stuck here. Presumably this part is related to (a), but what is the relation?

c. Show that $$\frac{(1+n)^n}{e^n} < n! < \frac{(1+n)^{1+n}}{e^n}$$ where $$n \in\mathbb Z$$ Hint: Let $k = 1,2,3,...n$ in part b)

Also stuck. Can induction be used here?

Answer 1

Hint. $f(x)=\frac{1}{x}$ is a decreasing function and thus $$ f(k+1)\le\int_{k}^{k+1}f(x)\,dx\le f(k), $$ To get the strict inequalities you need to do some more work.

Answer 2

We know that $y$ is positive and decreasing for $x>0$. For any continuous function, and any finite region of integration

$$ m ~(U-L) \le \int\limits_{L}^{U} \le M~(U-L) $$ where $m$ and $M$ are respectively the minimum and maximum value in the interval $[L,U]$

In the interval $[k, k+1]$, for $y=1/x$, $m = 1/(k+1)$ and $M=1/k$ and $U-L = 1$

Answer 3

I can only give you a hint as I am a bit too tired to work out the details. Hope this helps

$(1+1/k)^k$ is monotonically increasing and $(1+1/k)^{k+1}$ is monotonically decreasing and the limit of both is $e$. Hence (b)

(c) follows from (b) by induction. Consider the left inequality and suppose $$(1+n)^n < e^n n!$$ The induction step is $$(1+n+1)^{n+1} = \left(\frac{n+2}{n+1}\right)^{n+1} (n+1)^n (n+1) \lt\left(\frac{n+2}{n+1}\right)^{n+1} e^n n! \,(n+1) < e\, e^n (n+1)! $$ The last inequality follows from (b).