Using Lagrange's remainder, I have to prove that:

$\log(1+x) = \sum\limits_{n=1}^\infty (-1)^{n+1} \cdot \frac{x^n}{n}, \; \forall |x| < 1$

I am not quite sure how to do this. I started with the Taylor series for $x_0 = 0$:

$f(x_0) = \sum\limits_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!} \cdot x^n + r_n$, where $r_n$ is the remainder. Then, I used induction to prove that the n-th derivative of $\log(1+x)$ can be written as:

$f^{(n)} = (-1)^{n+1} \cdot \frac{(n-1)!}{(1+x)^n}, \forall n \in \mathbb{N}$

I plugged this formula into the Taylor series for $\log(1+x)$ and ended up with:

$f(x_0) = \sum\limits_{n=1}^\infty (-1)^{n+1} \cdot \frac{x^n}{n} + r_n$, which already looked quite promising.

As the formula which I have to prove doesn't have that remainder $r_n$, I tried to show that $\lim_{n \to \infty} r_n = 0$, using Lagrange's remainder formula (for $x_0 = 0$ and $|x| < 1$).

So now I basically showed that the formula was valid for $x \to x_0 = 0$. I also showed that the radius of convergence of this power series is $r = 1$, that is to say the power series converges $\forall |x| < 1$.

What is bugging me, is the fact, that to my opinion, the formula is only valid for $x \to 0$. I mean sure, the radius of convergence is 1, but does this actually tell me that the formula is valid within $(-1,1)$? I've never done something like this before, thus the insecurity. I'd be delighted, if someone could help me out and tell me, whether the things I've shown are already sufficient or whether I still need to prove something.

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    $\begingroup$ Robert Pollack has pointed out that my answer is wrong, so you may want to unaccept it. $\endgroup$ – Jonas Meyer Dec 10 '10 at 4:11
  • $\begingroup$ (But I have now added a sketch and reference for a correct method.) $\endgroup$ – Jonas Meyer Dec 10 '10 at 4:51

I think there is a problem with the above solution. In the estimate, $$|f^{(k+1)}(x)| \leq \left(\frac{1}{1-r}\right)^{k+1},$$ there is a dropped $k!$. Indeed, it should read, $$|f^{(k+1)}(x)| \leq \left(\frac{k!}{1-r}\right)^{k+1},$$ and thus $$ |r_k(x)| \leq \left( \frac{r}{1-r} \right)^{k+1} \cdot \frac{k!}{(k+1)!} = \left( \frac{r}{1-r} \right)^{k+1} \cdot \frac{1}{k+1}. $$ Unfortunately now this expression won't go to $0$ if $r>0$ (now the exponential term will dominate $\frac{1}{k+1}$).

The above solution does work for x in (-1/2,1). Here's a way to handle the remaining cases. In fact, let's just take x in (-1,0). Now Lagrange's form of the remainder gives: $$ r_k(x) = \int_0^x \frac{f^{(k+1)}(t)}{k!} (x-t)^{k} dt = \int_0^x \frac{(-1)^k}{(1+t)^{k+1}} (x-t)^{k} dt $$ Note that for $x<0$, the above integrand has the same sign for every $t$. In particular, $$ |r_k(x)| = \int_x^0 \frac{1}{(1+t)^{k+1}} (t-x)^{k} dt. $$ Consider $\frac{t-x}{1+t}$ as a function in $t$ with $x$ fixed. It then is an increasing function on [x,0] with maximal value of $-x$ when $t=0$. Thus, $$ |r_k(x)| \leq \int_x^0 (-x)^k \frac{1}{1+t} dt \leq \int_x^0 (-x)^k \frac{1}{1+x} dt = \frac{(-x)^{k+1}}{1+x}. $$ As desired, this last expression does go to $0$ as $k \to \infty$ as $-1<x<0$.

Note that in Spivak's Calculus, the stronger bound of $$ |r_k(x)| \leq \frac{(-x)^{k+1}}{(1+x)(1+k)} $$ is left as Exercise 16 in Chapter 20. I don't see how to get this, but I'm anxious that I'm just making some mistake in the computations above.

  • $\begingroup$ You're right, I made a serious error. It affects the answer in that the estimate I wanted to use will not work on $(-1,-\frac{1}{2})$. I'll update with the correction. $\endgroup$ – Jonas Meyer Dec 10 '10 at 4:04
  • $\begingroup$ Thanks for the correction. You added to your answer while I was updating mine, and I just noticed. The problem with the exponential term is when $r\gt\frac{1}{2}$, i.e., when $\frac{r}{1-r}\gt1$. $\endgroup$ – Jonas Meyer Dec 10 '10 at 4:44
  • $\begingroup$ Could you elaborate on the "similar argument"? Doesn't considering $\log(1-x)$ just put the hard part of the estimate on $(0,1)$? $\endgroup$ – Jonas Meyer Dec 10 '10 at 4:46
  • $\begingroup$ @Jonas: You're right. That "similar argument" seems to have no content at all. I'll think a bit to see if I actually had anything in mind. $\endgroup$ – Robert Pollack Dec 10 '10 at 15:48
  • $\begingroup$ @Jonas: I just put up an alternative solution. I couldn't figure out how to cross things out, so I just rewrote it all. $\endgroup$ – Robert Pollack Dec 10 '10 at 17:01

$f(x_0) = \sum\limits_{n=1}^\infty (-1)^{n+1} \cdot \frac{x^n}{n} + r_n$

That should say

$$f(x)=\sum_{n=1}^k (-1)^{n+1} \cdot \frac{x^n}{n} + r_k(x),$$

where $r_k$ is the error term of the $k^\text{th}$ partial sum. You want to use estimates to show that the error term goes to $0$ as $k$ goes to $\infty$, which will justify convergence of the series to $f(x)=\log(1+x)$.

Edit: I've struck through part of my answer that relied on a wrong estimate of the derivatives, as pointed out by Robert Pollack. With the missing $k!$ term, the estimate only works on $[-\frac{1}{2},1)$.

Added: To make this answer a little more useful, I decided to look up a correct method. Spivak in his book Calculus (3rd Edition, page 423) uses the formula $$\frac{1}{1+t}=1-t+t^2-\cdots+(-1)^{n-1}t^{n-1}+\frac{(-1)^nt^n}{1+t}$$ in order to write the remainder as $r_n(x)=(-1)^n\int_0^x\frac{t^n}{1+t}dt$. The estimate $\int_0^x\frac{t^n}{t+1}dt\leq\int_0^xt^ndt=\frac{x^{n+1}}{n+1}$ holds when $x\geq0$, and the harder estimate $\left|\int_0^x\frac{t^n}{1+t}\right|\leq\frac{|x|^{n+1}}{(1+x)(n+1)}$, when $-1\lt x\leq0$, is given as Problem 11 on page 430. Combining these, you can show that the sequence of remainders converges uniformly to $0$ on $[-r,1]$ for each $r\in(0,1)$.

Lagrange's form of the error term can be used to do this. The estimates, which follow from Taylor's theorem, are also found on Wikipedia. In this case, if $0\lt r\lt 1$, then $|f^{k+1}(x)|\leq \frac{1}{(1-r)^{k+1}}$ whenever $x\geq-r$, so you have the estimate $|r_k(x)|\leq \frac{r^{k+1}}{(1-r)^{k+1}}\frac{1}{(k+1)!}$ for all $x$ in $(-r,r)$, which you can show goes to $0$ (because (k+1)! grows faster than the exponential function $\left(\frac{r}{(1-r)}\right)^{k+1}$), thus showing that the series converges uniformly to $\log(1+x)$ on $(-r,r)$. Since $r$ was arbitrary, this shows that the series converges on $(-1,1)$, and the convergence is uniform on compact subintervals.

  • $\begingroup$ I don't follow. Why is it $f(x)$ and not $f(x_0)$? I mean sure, $x_0 = 0$ which is why it is no longer in the formula, but can you then simply write $f(x) = ...$? The rest I understand very well, but I don't understand how we arrive at a formula for $f(x)$ when we started with a Taylor series at $x_0 = 0$... $\endgroup$ – Huy Nov 27 '10 at 12:07
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    $\begingroup$ "$x_0=0$" just means that the series (and partial sums) are based at $0$, hence the powers of $x$, where in general there would be powers of $(x-x_0)$. If you are writing the series (or partial sums) of a function, you are concerned not just with its value at $x_0$, where all of the powers greater than $0$ vanish, making the convergence trivial. You want to know that the series for $f(x)$ converges for all $x$ in an interval, hence $x$ must be plugged in, not just the particular value $x_0$. The Wikipedia article I linked to has more on this. $\endgroup$ – Jonas Meyer Nov 27 '10 at 12:12
  • $\begingroup$ Just to be clear, already my start is "wrong", it should be f(x) instead of f(x0) and it should have two partial sums (instead of an infinite sum and the remainder with an "undefined" index n)? If so, I think I understood your explanation and I want to thank you a lot for it. One last question: Does the radius of convergence always tell me, that for all x in the radius, the power series converges to the function (e.g. log(1+x)) or does it only tell me that for all x in the radius, the power series converges (to some value)? Because up to now I assumed the latter to be correct... $\endgroup$ – Huy Nov 27 '10 at 12:27
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    $\begingroup$ @Huy: That is correct, at the start it should be $f(x)$. And the remainder term is for when you have a partial sum, and you want to show that it goes to $0$ as $k\to\infty$ so that it is no longer there when you take the series. You're welcome. For your last question, as long as you can show that the remainder goes to zero (and it always does if you're within the radius of convergence of an analytic function, i.e. a function with power series expansion on a given interval), then the power series converges to the function, not just to any old value. $\endgroup$ – Jonas Meyer Nov 27 '10 at 12:32
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    $\begingroup$ That is why you have to show that the remainder term goes to zero, not just that the series converges. There are functions whose Taylor series converge but do not converge to the function (an example is $g(x)=e^{-1/x^2}$ if $x\neq0$, $g(0)=0$). If you checked the remainder term of such functions, they would therefore not converge to $0$. $\endgroup$ – Jonas Meyer Nov 27 '10 at 12:34

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