# Taylor expansion of ln(1+x)

• May 3rd 2010, 07:35 PM
Pinkk
Taylor expansion of ln(1+x)
Okay, so the question is regarding the relation between the interval of convergence and the interval for which $f(x) = \ln (1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}x^{n}$. I know they are the same interval, but I am trying to show this by showing that the remainder function, $R_{n}$ tends to $0$ only when $x\in (-1,1]$. So I have to show that $\frac{f^{(n)}(y)}{n!}x^{n}$ tends to $0$ where $y$ lies between $0$ and $x$ (if this approach is wrong, please correct me). It can be observed that $f^{(n)}(x)=(-1)^{n+1}(n-1)!(1+x)^{-n}$ so plugging that in I have to show that $\frac{1}{n}|\frac{x}{1+y}|^{n}$ tends to $0$ (because if $|R_{n}|$ tends to $0$, then so does $R_{n}$). So the condition for that to tend to zero is that $|\frac{x}{1+y}| \le 1$. And now this is where I get stuck showing that $x$ must lie in $(-1,1]$.

A secondary question is why does this still converge on some interval even though on that interval, not all the derivatives $f^{(n)}(x)$ are not bounded by some single constant $C$ on that interval?
• May 3rd 2010, 07:51 PM
Do you know that an alternating series converges if its terms go to zero in magnitude?
• May 3rd 2010, 07:56 PM
Pinkk
Yes, that part I have about the radius of convergence for the series. I am trying to show that the same interval of convergence is the interval for which the Taylor series represents the function, and for that I believe I have to show that $R_{n}$ tends to $0$.

Another follow up question would be why is there this relationship between the interval of convergence of the series and the interval for which the function is equal to the Taylor expansion. Is there a generalized proof that shows that the two intervals must always be equal?
• May 3rd 2010, 08:17 PM
So intervals of convergence do not imply interval for which the function is analytic. Is there a proof for why in this case of $\ln (1+x)$ the two intervals are the same? Right now I'm just trying to show that they are the same (which I still need help with proving if $x\notin (-1,1]$ that there is always a case where there exists $y$ between $0$ and $x$ where $|y+1| > |x|$) but is there a proof where one interval implies the other for this particular function?
Okay, even if I use Cauchy's, I think I'll still run into the same problem that I need to show there's always a case where if $x > 1$ there will be a $y$ such that $x - y > 1$.
Maybe try Bernoulli's inequality: for nonzero y>-1 and integers n>1, $(1+y)^n>1+ny$.