How to justify this estimation?

Question

I was looking (again) at the series $$\sum_{n = 0}^{+\infty} \frac{3^n}{n!}$$ to give an estimation of its sum withouth the knowledge of the exponential series, and following the steps of a majorisation I faced a dark point.

The steps are these: consider

$$\frac{3^n}{n!} = \frac{3\cdot 3\cdot 3\cdot 3 \ldots}{1\cdot 2\cdot 3\cdot \ldots n} = \frac{3\cdot 3\cdot 3}{1\cdot 2\cdot 3} \frac{3^{n-3}}{4\cdot 5\ldots n} = \frac{9}{2}\frac{3^{n-3}}{4\cdot 5\ldots n} $$

Now we can say that $\forall n \geq 3$ it holds $n! \geq 4^{n-3}$ hence $$\frac{1}{n!} \leq \frac{1}{4^{n-3}}$$

I understood this "trick" has been done because in this way we can use the Geometric series (which we can), indeed then:

$$\frac{3^n}{n!} \leq \frac{9}{2} \left(\frac{3^{n-3}}{4^{n-3}}\right)^n$$

$$\frac{3^n}{n!} \leq \frac{9}{2} \frac{4^3}{3^3} \left(\frac{3}{4}\right)^n$$

In this was I can write

$$\sum_{n = 3}^{+\infty} \frac{3^n}{n!} \frac{32}{3} \leq \sum_{n = 3}^{+\infty}\left(\frac{3}{4}\right)^n$$

and so on...

• Question is about this part:
  $$\frac{3^n}{n!} \leq \frac{9}{2} \left(\frac{3^{n-3}}{4^{n-3}}\right)$$

How can we justify this? If I read this, what I read is: separately that $\frac{1}{n!} \leq \frac{1}{4^{n-3}}$ and $3^n \leq \frac{9}{2}3^{n-3}$ that is $3^n \leq \frac{3^{n-1}}{2}$ which is not true.

I cannot understand this part.

Thank you!

Answer 1

First of all, there is no $n$ exponent after the parenthesis. Then write explicitly the first three terms: $$\frac{3^n}{n!}=\frac{3\cdot3\cdot3\cdot 3^{n-3}}{1\cdot2\cdot3\cdot 4\cdot ...\cdot n}\le\frac{3\cdot3\cdot3}{1\cdot 2\cdot 3}\frac{3^{n-3}}{4\cdot4\cdot ...\cdot 4}=\frac92\frac{3^{n-3}}{4^{n-3}}=\frac92\left(\frac34\right)^{n-3}$$

Answer 2

The trick can be employed like this:

$\frac{1}{n!}\leq\frac{1}{4^{n-3}}$ and $\frac{3^n}{n!}\leq \frac{3^n}{4^{n-3}}=\frac{3^33^{n-3}}{4^{n-3}}=3^3(\frac{3^{n-3}}{4^{n-3}})=\frac{3^34^3}{3^3}(\frac{3}{4})^n=4^3(\frac{3}{4})^n$, for $n\geq3$.

Answer 3

For your curiosity

Make the problem more general and think about the complete and incomplete gamma functions $$S_p(a)=\sum_{n = 0}^p \frac{a^n}{n!}=e^a \,\,\frac{\Gamma (p+1,a)}{\Gamma (p+1)}$$ and (have a look here) its asymptotic is $$S_p(a)=e^a-\frac{ a^{p+1}\, e^p}{p^{p+\frac{3}{2}}\sqrt{2 \pi} }-\frac {(12a-13)\, a^{p+1}\, e^p}{12p^{p+\frac{5}{2}}\sqrt{2 \pi} }+\cdots$$

Using the first term $$\log \big( e^a-2 \sqrt{\pi }S_p(a)\big)=p+(1+p)\log(a)-\left(p+\frac{3}{2}\right) \log (p)$$

For $a=3$, the relative error is smaller than $1.00$% if $p>5$, smaller than $0.10$% if $p>7$, smaller than $0.01$% if $p>9$.