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

[jegues]

Homework Statement

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

Homework Equations

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 a lot 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, let's 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!

 

[jegues]

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!