Is sequence 2^n/n increasing or decreasing?

[Dethrone]

I'm starting sequences and series, and I have to prove that the following is increasing, nonincreasing, decreasing, or nondecreasing.

$$S_n = \frac{2^n}{n!}$$

I've tried $S_{n+1}-S_n >$ or $<$ than $0$, $\frac{S_{n+1}}{S_n}$, and I don't think I've learned differentiating factorials so I won't even attempt that. Any hints?

 

[MarkFL]

Rather than continuous differentiation, I would look at discrete differencing. :D

 

[Dethrone]

What do you mean by discrete differencing?

 

[MarkFL]

Suppose we define:

\(\displaystyle S_n\equiv\frac{a_n}{b_n}\)

where:

\(\displaystyle a_n=2^n,\,b_n=n!\)

And so we could observe that $n$ is not continuous, but is presumably the natural numbers, so look at:

\(\displaystyle \frac{\Delta a_n}{\Delta n}=\frac{a_{n+1}-a_{n}}{n+1-n}=2^{n+1}-2^n\)

and

\(\displaystyle \frac{\Delta b_n}{\Delta n}=\frac{b_{n+1}-b_{n}}{n+1-n}=(n+1)!-n!\)

What is the ratio of the differences?

 

[Prove It]

What do you mean by discrete differencing?

Notice that $\displaystyle \begin{align*} 2^n = 2 \cdot 2 \cdot 2 \cdot \dots \cdot 2 \end{align*}$ (n times), while $\displaystyle \begin{align*} n! = n \cdot (n - 1) \cdot (n - 2) \cdot \dots \cdot 3 \cdot 2 \cdot 1 \end{align*}$.

Since it is clear that $\displaystyle \begin{align*} n! >> 2^n \end{align*}$, that means that the bottom grows much faster than the top.

Thus the limit of the sequence must be 0 and so the sequence must be decreasing.

 

[Dethrone]

The ratio of the differences is
$$\frac{2^{n+1}-2^n}{(n+1)!-n!}$$
but I'm not sure how I can simplify the denominator:
$$\frac{2^n}{(n+1)!-n!}$$

 

[MarkFL]

The ratio of the differences is
$$\frac{2^{n+1}-2^n}{(n+1)!-n!}$$
but I'm not sure how I can simplify the denominator:
$$\frac{2^n}{(n+1)!-n!}$$

Hint:

\(\displaystyle (n+1)!=(n+1)n!\)

 

[Dethrone]

Notice that $\displaystyle \begin{align*} 2^n = 2 \cdot 2 \cdot 2 \cdot \dots \cdot 2 \end{align*}$ (n times), while $\displaystyle \begin{align*} n! = n \cdot (n - 1) \cdot (n - 2) \cdot \dots \cdot 3 \cdot 2 \cdot 1 \end{align*}$.

Since it is clear that $\displaystyle \begin{align*} n! >> 2^n \end{align*}$, that means that the bottom grows much faster than the top.

Thus the limit of the sequence must be 0 and so the sequence must be decreasing.

I thought about that too, but I thought there was a more mathematical way of presenting it.@Mark,

$$\frac{2^n}{n!(n)}$$

Are we suppose to conclude now that for discrete differences, the denominator increases much faster than the numerator, meaning that sequence must be decreasing? Or is that statement premature?

 

[Prove It]

I thought about that too, but I thought there was a more mathematical way of presenting it.@Mark,

$$\frac{2^n}{n!(n)}$$

Are we suppose to conclude now that for discrete differences, the denominator increases much faster than the numerator, meaning that sequence must be decreasing? Or is that statement premature?

To use English to explain logical conclusions IS mathematical!

 

[I like Serena]

My suggestion would be to write out elements of the sequence until a pattern becomes clear. About 5 or 6 of them should do. Then you should also be able to say why (plain English will do).

 

[MarkFL]

I thought about that too, but I thought there was a more mathematical way of presenting it.@Mark,

$$\frac{2^n}{n!(n)}$$

Are we suppose to conclude now that for discrete differences, the denominator increases much faster than the numerator, meaning that sequence must be decreasing? Or is that statement premature?

Prove It's suggestion is a good way to look at it. Simple and neat.

Perhaps a better approach than the one I initially suggested would be to look at:

\(\displaystyle \frac{\Delta S_n }{\Delta n}\)

 

[Dethrone]

ILY, I thought that would work too, but believed there must have been a better, less mundane way of doing it.

Also, for completeness, I say that the sequence is nonincreasing because $S_1 = S_2 = 2$ and it is decreasing only when $n \ge 2$