Proving the Limit in a Power Series: How Does the Solution Work?

[ecoo]

Homework Statement

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.

Homework Equations

Convergent power series form

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?

 

[Mark44]

Homework Statement

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.

Homework Equations

Convergent power series form

The Attempt at a Solution

I attempted to represent the series with a Maclaurin series, but that leads nowhere.

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

Could someone help explain why the solution given works?

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

 

[ecoo]

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?

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.

 

[jbunniii]

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.

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 $\lim_{n\to\infty}1/n = 0$.

 

[ecoo]

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 $\lim_{n\to\infty}1/n = 0$.

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

 

[Mark44]

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

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.

 

[ecoo]

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.

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

 

[Mark44]

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

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.

 

[ecoo]

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.

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

 

[Mark44]

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.

 

[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 $a_0$.

 

[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.