2

I was working with the Maclaurin expansion of $f(x) = \ln(1+x)$. I will present the expansion quickly to get to the point I had trouble with.

$I$. $f(x) = \ln(1+x)$ is such that

$$\{ f^{(n)}(x) \}_{n=1}^\infty = \{\frac{1}{1+x}, -\frac{1}{(1+x)^2}, \frac{2}{(1+x)^3}, - \frac{6}{(1+x)^4},..., (-1)^{n-1}\frac{(n-1)!}{(1+x)^n} \}$$

$II$. From $I$ it follows that

$$\{ f^{(n)}(0) \}_{n=1}^\infty = \{1, -1, 2, - 6,..., (-1)^{n-1}(n-1)! \}$$

$III$. Then

$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n=\sum_{n=1}^\infty (-1)^{n-1}\frac{x^n}{n}$$

So far so good. But I was now requested to show for what $x$ the series is equal to the $f$. I attempted to use Taylor's inequality, asking whether

$$|f^{(n+1)}(x)| = \frac{n!}{(1+x)^{n+1}} \leq M$$

when $x$ is bounded around $0$ by some distance $d$ (equivalently, for $|x| \leq d$). However, I found it difficult to think of values of $x$ for which $\frac{n!}{(1+x)^{n+1}}$ is bounded under some $M$. At this point, I did not know how to proceed. How should I go about this?

lafinur
  • 3,322
  • $\limsup_n \sqrt[n]{{1 \over n}} = 1$. – copper.hat Sep 22 '22 at 17:07
  • @copper.hat ... that shows radius of convergence of the series is $1$, but doesn't show that it converges to $\log(1+x)$. – GEdgar Sep 22 '22 at 17:09
  • My approach would be: start with $f'(x) = 1/(1+x)$, where its Maclaurin series is a geometric series, and presumably you know all about that convergence, and what it converges to. Then isn't there a theorem about integrating a power series term-by-term? The only thing left will be to determine what happens at the endpoints $x=\pm 1$. – GEdgar Sep 22 '22 at 17:11

1 Answers1

1

In the Maclaurin seriesthe term with $x^n$ is actually

$\dfrac{f^{(n)}(0)}{n!}.$

So the factor of $n!$ you see in the derivative is canceled by the denominator, and thereby you have bounded terms when $|x|\le1$.

Note, however, that bounded terms are only necessary not sufficient fir convergence of the series. You have to check a little more carefully to find the precise convergence region.

We can develop a series for the logarithm with rational functions that converges for all positive arguments. See this question.

Oscar Lanzi
  • 39,403