1. ## Taylor remainder

I was finding the Taylor expansion for $\displaystyle f(x)=\ln x$ at $\displaystyle x=1$ and I've found that $\displaystyle f^{(n)}(x)=\frac{(-1)^{n+1}(n-1)!}{x^n}$ but I need to show that the remainder tends to zero as $\displaystyle n\to\infty,$ my problem is that I don't know what's the form of the remainder and how to show that it tends to zero.

2. Originally Posted by Astrid
I was finding the Taylor expansion for $\displaystyle f(x)=\ln x$ at $\displaystyle x=1$ and I've found that $\displaystyle f^{(n)}(x)=\frac{(-1)^{n+1}(n-1)!}{x^n}$ but I need to show that the remainder tends to zero as $\displaystyle n\to\infty,$ my problem is that I don't know what's the form of the remainder and how to show that it tends to zero.

I think that as you tend to infinity $\displaystyle (-1)^{n+1}$ makes no difference it just changes the sign so igrore this, then you have $\displaystyle \frac{(n-1)!}{x^n}$ as you know that $\displaystyle x^{n}$ tends to infinite quicker than (n-1)! tends to infinit you get $\displaystyle \frac{small infinite}{big infinite}$

hope this helps

3. no, it doesn't, sorry.

what's the formula of the remainder? how to compute it's limit?

4. Originally Posted by Astrid
no, it doesn't, sorry.

what's the formula of the remainder? how to compute it's limit?

Okay if u think about any fraction if u have a large number on the bottem and a small number on the top you end up with ur problem tending to 0.
N sorry dont know what the formula for the remainder is.

bobisback