# Taylors Series/Remainder for ln(x) about x = 1

## Homework Statement

Determine whether the Taylors Series for ln(x) converges to ln(x) around x=1, for x [1/2, 2].

## The Attempt at a Solution

$$f(x) = ln(x)$$

$$f^{1}(x) = \frac{1}{x}$$

$$f^{2}(x) = \frac{-1}{x^{2}}$$

$$f^{3}(x) = \frac{2}{x^{3}}$$

$$f^{4}(x) = \frac{-6}{x^{4}}$$

$$\vdots$$

$$f^{n}(x) = \frac{(-1)^{n-1}(n-1)!}{x^{n}}$$

So,

$$ln(x) = (x-1) - \frac{1}{2}(x-1)^{2} + \frac{1}{3}(x-1)^{3} - \cdots + \frac{(-1)^{n-1}}{n}(x-1)^{n} + R_{n}(1,x)$$

Where,

$$R_{n}(1,x) = \frac{(-1)^{n}}{(n+1)} \cdot \left( \frac{(x-1)}{z_{n}} \right)^{n+1}$$

We can see that the second term (The one with $$z_{n}$$ in it) is the one that causes problems. It is trivial to show that the limit as n goes to infinity of the first term is 0. (We proved it in class)

So there are 3 cases that we should consider:

Case 1: x=1, everything will converge to 0.

Case 2: $$\frac{1}{2} \leq x < 1$$

It follows then that,

$$\frac{1}{2} \leq x < z_{n} <1$$

We can use, $$\frac{1}{2} \leq x < 1$$ to try to gain for information about the term that is causing problems with our remainder.

$$\frac{1}{2} \leq x < 1 \Rightarrow \frac{1}{2} - 1 \leq x-1 < 1-1 \Rightarrow \frac{-1}{2} \leq x-1 < 0 \Rightarrow \frac{-1}{2z_{n}} \leq \frac{x-1}{z_{n}} < 0$$

Looking back at,

$$\frac{1}{2} \leq x < z_{n}$$

we can see that,

$$z_{n} \geq \frac{1}{2}$$ or $$2z_{n} \geq 1$$

Working with this some more we can see that,

$$\frac{1}{2z_{n}} \leq 1 \Rightarrow \frac{-1}{2z_{n}} \geq -1$$

Therefore we can say that,

$$-1 \leq \frac{-1}{2z_{n}} < \frac{x-1}{z_{n}} < 0$$

i.e. $$-1 < \frac{x-1}{z_{n}} < 0 \Rightarrow |\frac{x-1}{z_{n}}| < 1$$

This is what our professor wrote down in order to prove the following,

$$|R_{n}(1,x)| = \left| \frac{(-1)^{n}}{n+1} \cdot \frac{(x-1)^{n+1}}{z_{n}} \right| < \frac{1}{n+1} \cdot (1)$$

Therefore,

$$lim_{n \rightarrow \infty} |R_{n}(1,x)| = 0 \Rightarrow lim_{n \rightarrow \infty} R_{n}(1,x) = 0$$.

I'm confused about how he got rid of that unwanted term,

$$\left(\frac{(x-1)}{z_{n}}\right)^{n+1}$$

by using the inequalities stated above. He seems to jump around alot and it's hard to follow.

EDIT: Looking at it again, he first shows that,

$$\frac{x-1}{z_{n}} < 0$$

Now if he can prove that this is also larger than -1 it will be a small negative number inbetween (0, -1) and no matter what power we raise it to, lets say n+1, it will still be really close to 0.

Is this his attack?

Can someone try to explain his reasoning another way or offer me some insight? If I tried to do this problem without looking at my notes I don't think I'd be able to reasonably conclude that,

$$lim_{n \rightarrow \infty} R_{n}(1,x) = 0$$

For Case 2: $$\frac{1}{2} \leq x < 1$$

Any advice/tips/ideas/insight would be greatly appreciated! (After I will attempt case 3)

Thanks again!

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Case 3: $$1 < x \leq 2$$ seems to make some more sense so I'll attempt it in this post.

It follows that,

$$1 < z_{n} < x \leq 2$$

Again, we can use, $$1 < x \leq 2$$ to gain more information about our problematic term in $$R_{n}(1,x)$$.

One can see that,

$$0 < x-1 \leq 1$$

$$0 < \frac{x-1}{z_{n}} \leq \frac{1}{z_{n}}$$

From here we can see that,

$$\frac{1}{z_{n}} < 1$$

So here's where I'm at,

$$0 \leq |R_{n}(1,x)| < ?$$

We need to figure out an upper bound so we can apply squeeze theorem.

We know that,

$$\frac{x-1}{z_{n}}$$ is less than 1 but larger than 0.

Hmm...

I'm not sure how I can find that upper bound I've been looking for. If I think of anything I'll simply edit this post.

Feel free to make any suggestions if you have any!