How do you solve an inequality that involves factorials?

Question

I have this inequality that's part of a larger problem (regarding Taylor expansions). I have the steps to the solution, but I don't understand this part:

Step 1: $e/1000 > 1/(n+1)!$

Step 2: $1000/e < (n+1)!$

Step 3: $(n+1) = 6$

I'm lost on how that's true? How do you go from step 2 to step 3?

Answer 1

In general, you just use a calculator (or a table of values) to try small values of $n$ and see when the inequality is satisfied. For example, using a calculator, we know that $\dfrac{1000}{e} \approx 367.88$. Also, it is pretty easy to verify that \begin{align} 5! = 120 < \dfrac{1000}{e} < 720 = 6! \end{align} Hence, we choose $(n+1) = 6$; in other words $n=5$ is the smallest integer value of $n$ which satisfies the inequality \begin{align} \dfrac{1000}{e} > \dfrac{1}{(n+1)!} \end{align}

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In fact any positive integer $n \geq 5$ also satisfies the inequality above, so strictly speaking, we should say $n \geq 5$, not just $n=5$

Answer 2

Welcome to Maths SX! Everyone knows (at least) that $5!=120$, hence $6!=720$. On the other hand, $2<\mathrm e <4$, so $$\frac{1000}4=250<\frac{1000}{\mathrm e}<\frac{1000}2=500,$$ so the first factorial greater than $\frac{1000}{\mathrm e}$ is indeed $6!$.