Prove an inequality relating sums and integrals

Prove that

$\sum_{k=1}^{n-1} f(k) \leq \int_1^n f(x) \, dx \leq \sum_{k=2}^n f(k)$

where $f$ is a nonnegative, increasing function defined for all $x \geq 1$ .

Use this to deduce the inequalities

$e n^n e^{-n} < n! < e n^{n+1} e^{-n}$

by taking $f(x) = \log x$ .

Proof. Since $f(x)$ is increasing we know $f([x]) \leq f(x) \leq f([x+1])$ (where $[x]$ denotes the greatest integer less than or equal to $x$ ) for all $x \geq 1$ . Define step functions $s$ and $g$ by

$s(x) = f([x]), \qquad t(x) = f([x+1]).$

Then $s$ and $t$ are constant on the open subintervals of the partition

$P = \{ 0,1,2, \ldots, n \}.$

So,

$\begin{align*} \int_1^n s(x) \, dx = \sum_{k=1}^{n-1} s_k (x_k - x_{k-1}) = \sum_{k=1}^{n-1} f(k) \\[9pt] \int_1^n t(x) \, dx = \sum_{k=1}^{n-1} t_k (x_k - x_{k-1}) = \sum_{k=1}^{n-1} f(k+1) = \sum_{k=2}^n f(k). \end{align*}$

Since $f$ is integrable we must have

$\int_1^n s(x) \, dx \leq \int_1^n f(x) \, dx \leq \int_1^n t(x) \, dx$

for every pair of step functions $s \leq f \leq t$ . Hence,

$\sum_{k=1}^{n-1} f(k) \leq \int_1^n f(x) \, dx \leq \sum_{k=2}^n f(k). \qquad \blacksquare$

Next, if we take $f(x) = \log x$ (which is nonnegative and increasing on $x \geq 1$ ) we have

$\begin{align*} &&\sum_{k=1}^{n-1} \log k \leq \int_1^n \log x \, dx \leq \sum_{k=2}^n \log k \\[9pt] \implies && \log(n-1)! \leq n \log n - n + 1 \leq \log n! \\[9pt] \implies && (n-1)! \leq e n^n e^{-n} \leq n!. \end{align*}$

From this we then get two inequalities

$n! \leq en^{n+1} e^{-n} \quad \text{and} \quad e n^n e^{-n} \leq n!.$

Therefore,

$en^n e^{-n} \leq n! \leq e n^{n+1} e^{-n}.$

Stumbling Robot

Prove an inequality relating sums and integrals

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