Binomial Theorem related proofs

[h.shin]

Homework Statement

Let a be a fixed positive rational number. Choose (and fix) a natural number M>a.
Use (a^n)/(n!)[itex]\leq[/itex](a^M/(M!))(a/M)^(n-M) to show that, given e>0, there exists an N[itex]\in[/itex][itex]N[/itex] such that for all n[itex]\geq[/itex]N, (a^n)/n! < e.

Homework Equations

The Attempt at a Solution

In a previous problem, I saw that when M>n then (a^n)/(n!)<(a^M/(M!))(a/M)^(n-M). So I thought i could maybe use that to come up with a N in relations to e. but I'm not so sure how to do this. I know the equations are long and ugly, but please help.

 

[mighty2000]

Thank you for sharing your thoughts and attempt at a solution. Let's work through this problem together.

First, let's review the given inequality: (a^n)/(n!)<=(a^M/(M!))(a/M)^(n-M).

We know that a and M are fixed positive rational numbers, and since we are trying to prove that (a^n)/n! < e for all n≥N, we can choose a specific value for N. Let's say N=M, since we know that for n>M, the right side of the inequality will become smaller and smaller as n increases.

Now, let's substitute N=M into the inequality: (a^n)/(n!)<=(a^M/(M!))(a/M)^(n-M) becomes (a^n)/(n!)<=(a^M/(M!))(a/M)^(M-M). Simplifying further, we get (a^n)/(n!)<=(a^M/(M!)).

We can rewrite this as (a^n)/(n!)<=(a^M)/(M!)*(1/M)^n.

Since we are trying to prove that (a^n)/n! < e for all n≥N, we can choose a specific value for e. Let's say e=(a^M)/(M!)*(1/M)^N. This means that for all n≥N, (a^n)/n! < (a^M)/(M!)*(1/M)^N.

To show that (a^n)/n! < e, we need to show that (a^n)/n! < (a^M)/(M!)*(1/M)^N < (a^M)/(M!)*(1/M)^n.

We can rewrite this as (a^n)/(n!) < (a^M)/(M!)*(1/M)^(n-N).

Now, let's choose a specific value for N, such as N=2M. This means that for all n≥2M, (a^n)/n! < (a^M)/(M!)*(1/M)^(n-2M).

We can rewrite this as (a^n)/n! < (a^M)/(M!)*(1/M)^n.

Since we chose a specific value for e and N, we have shown that for all n≥N, (a^n)/n! < e. Therefore, the statement is proven.

I hope