Taylor series for logarithm converges towards logarithm

by Ulrik   Last Updated June 12, 2019 08:20 AM

Is there a way to show that the Taylor series around 0 of $f(x) = \ln(1-x)$ converges towards $f$ on the interval $(-1,1)$, just by considering the remainder from the Taylor polynomial? I'm having a little trouble with this.

The series is

$ T_n(x) = - \sum_{k=1}^n \frac{x^k}{k}$

The convergence on $(-1,0]$ is not a problem, but on $[0,1)$ things start to get a little complicated. Let $0<x<1$. then $|f^{(n+1)}(t)| = \frac{n!}{(1-t)^{n+1}} \leq \frac{n!}{(1-x)^{n+1}}$ for all $t \in [0,x]$. The Lagrange form of the remainder then tells us that

$|R_n(x)| \leq \frac{n!}{(1-x)^{n+1} (n+1)!}x^{n+1} = \frac{1}{n+1} \left( \frac{x}{1-x} \right)^n$

But if $x > 1/2$, this goes to infinity. I have tried to use the integral form of the remainder as well, but with no luck.

Using the machinery of power series, it's easy to prove the convergence. But is there a way using only theory about Taylor polynomials?



Answers 1


The only proof of this that I know of uses the Cauchy form of the remainder. If $f^{(n+1)}(x)$ exists for all $x$ between $0$ and $h$ and is continuous on this region$^\dagger$, then this form of the remainder can be shown to hold by using the integral form of the remainder: $$R_{n+1}(x)= {1\over n!} \int_0^x f^{(n+1)}(t)(x-t)^n\,dt $$ and applying the Second Mean Value Theorem for Integrals to this expression (with $g(t)=1$).


So, let's use the Cauchy form of the remainder to show that the Taylor (Maclaurin) series of $\ln(1+x)$ converges to $\ln(1+x)$ for $-1<x<0$:

Let $f(x)=\ln(1+x)$. Using the Cauchy form of the remainder, one has $$ \ln(1+x) =x-{x^2\over2}+\cdots+{(-1)^{n-1} x^n\over n} + R_{n+1}(x), $$ where $$ R_{n+1}(x)={f^{(n+1)}(c)\over n!}(x-c)^n x $$ for some $c$ between $0$ and $x$. Note we may write $$ R_{n+1}(x)= {f^{(n+1)}(\theta x)\over n!}(1-\theta)^n x^{n+1} $$ for some $0\le\theta\le1$.

Evaluating the required derivative, we have $$ R_{n+1}(x)=(-1)^n{x^{n+1}(1-\theta)^n\over (1+\theta x)^{n+1}}. $$

Now, for $-1<x<0$, note that $$ 1+\theta x\ge 1+x $$ and $$ 0\le{1-\theta\over 1+\theta x}\le 1; $$ whence $$ | R_{n+1}(x)| =\biggl|{x^{n+1}(1-\theta)^n\over (1+\theta x)^{n+1}} \biggr| =\biggl|{1-\theta\over 1+\theta x} \biggr|^n \cdot{|x|^{n+1}\over |1+\theta x|} \le{|x|^{n+1}\over 1+x}\ \buildrel{n\rightarrow\infty}\over\longrightarrow\ 0. $$


$^\dagger$ The continuity of $f^{(n+1)}$ is not required here; but in this case, a different proof is required.

David Mitra
David Mitra
February 18, 2013 14:26 PM

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