# Proving a limit (x^n/(n!))

1. May 29, 2016

### ecoo

1. The problem statement, all variables and given/known data

Section is on using power series to calculate functions, the problem is on proving the limit, solution is also attached but I do not see how the solution proves the limit.

2. Relevant equations

Convergent power series form

3. The attempt at a solution

I attempted to represent the series with a Maclaurin series, but that leads nowhere. Could someone help explain why the solution given works?

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2. May 29, 2016

### Staff: Mentor

The problem doesn't ask you to find a Maclaurin series -- it just asks you to prove the limit that is given.

The proof seems very straightforward to me. What part of it do you not understand?

3. May 29, 2016

### ecoo

I don't see how showing that the series converges for all x proves that the limit as n approaches 0 is equal to 0 for all x.

4. May 29, 2016

### jbunniii

This is true in general for any convergent series. If $\sum_{n=0}^{\infty}a_n$ converges, then $\lim_{n \to \infty} a_n = 0$. This is sometimes called the n'th term test when stated in its contrapositive form: if the sequence $a_n$ does not converge to zero, then the series $\sum_{n=0}^{\infty} a_n$ diverges.

The converse is false, as shown by the example $\sum_{n=1}^{\infty}(1/n)$, which diverges even though $\lim_{n\to\infty}1/n = 0$.

5. May 30, 2016

### ecoo

That's true, but aren't we trying to prove the limit as n approaches 0, not infinity?

6. May 30, 2016

### Staff: Mentor

What you just wrote makes no sense. They are trying to prove that the limit is zero, as n approaches infinity. More specifically, that $\lim_{n \to \infty}\frac {x^n}{n!} = 0$. They are using the Ratio Test to prove this.

In addition, this work also proves that the series $\sum_{n = 0}^{\infty}\frac{x^n}{n!}$ is convergent, for the reason jbunniii gave.

7. May 30, 2016

### ecoo

The first sentence, they are trying to prove the limit is 0 as n approaches 0.

8. May 30, 2016

### Staff: Mentor

You're right. The type is so small in the image you posted that I misread the 0 for ∞.

I'm pretty sure that the $n \to 0$ under "lim" is a typo. The other work shown suggests strongly that it should be $\lim_{n \to \infty}$ which is what I mistook it to be.

9. May 30, 2016

### ecoo

Ah, I see. How would one go about prove the limit as n approaches 0, or would it be very hard?

10. May 30, 2016

### Staff: Mentor

No, it's not hard. If $x \ne 0$, $\lim_{x \to 0}\frac {x^n}{n!}$ = 1. As $n \to 0, x^n \to 1$, and 0! is defined to be 1. $0^0$ is indeterminate, though, which is why I qualified what I said about x not being equal to zero.

11. May 30, 2016

### jbunniii

It doesn't make much sense to talk about the limit as $n$ approaches zero. This is because $n$ is an integer, so the only way it can approach zero is by being zero. So, the limit of a sequence $a_n$ as $n$ approaches zero is simply $a_0$.

12. Jun 5, 2016

### Eclair_de_XII

Perhaps try separating the $|x|$ and the $\frac{1}{n+1}$ and you may see why it works for any real x.

It has to do with the value of $\frac{1}{n+1}$ as n approaches infinity.