Compute the taylor series of $\ln(1+x)$

I've first computed derivatives (up to the 4th) of ln(1+x)

$f^{'}(x)$ = $\frac{1}{1+x}$
$f^{''}(x) = \frac{-1}{(1+x)^2}$
$f^{'''}(x) = \frac{2}{(1+x)^3}$
$f^{''''}(x) = \frac{-6}{(1+x)^4}$
Therefore the series:
$\ln(1+x) = f(a) + \frac{1}{1+a}\frac{x-a}{1!} - \frac{1}{(1+a)^2}\frac{(x-a)^2}{2!} + \frac{2}{(1+a)^3}\frac{(x-a)^3}{3!} - \frac{6}{(1+a)^4}\frac{(x-a)^4}{4!} + ...$

But this doesn't seem to be correct. Can anyone please explain why this doesn't work?

The supposed correct answer is:
As $\ln(1+x) = \int (\frac{1}{1+x})dx$
$\ln(1+x) = \Sigma_{k=0}^{\infty} \int (-x)^k dx$

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    I should say here that the Taylor (or Maclaurin) series $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$ converges [if and only if (iff)](https://en.wikipedia.org/wiki/If_and_only_if) $-1 – user236182 Sep 17 '17 at 01:18

4 Answers4


You got the general expansion about $x=a$. Here we are intended to take $a=0$. That is, we are finding the Maclaurin series of $\ln(1+x)$. That will simplify your expression considerably. Note also that $\frac{(n-1)!}{n!}=\frac{1}{n}$.

The approach in the suggested solution also works. We note that $$\frac{1}{1+t}=1-t+t^2-t^3+\cdots\tag{1}$$ if $|t|\lt 1$ (infinite geometric series). Then we note that $$\ln(1+x)=\int_0^x \frac{1}{1+t}\,dt.$$ Then we integrate the right-hand side of (1) term by term. We get $$\ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots,$$ precisely the same thing as what one gets by putting $a=0$ in your expression.

Alex Ortiz
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André Nicolas
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    If a=0 in my expansion, $ln(1+x) = 0 + x - \frac{x^2}{2!} + \frac{2(x^3)}{3!} - \frac{6(x^4)}{4!}+...$ Which is not the same ! – Minu Jul 26 '14 at 00:20
  • Why doesn't this method work? – Minu Jul 26 '14 at 00:23
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    It is the same, I mentioned the cancellation in the answer. For instance, $2!/3!=1/3$, $6/4!=1/4$, and so on. – André Nicolas Jul 26 '14 at 00:24
  • aren't the signs different? – Minu Jul 26 '14 at 00:27
  • In my expansion, $ln(1+x)= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$ Which is not the same, Is it? – Minu Jul 26 '14 at 00:31
  • Since $\ln (1+a)=0$ if $a=0$, the series you got is $x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$. Your signs are right. – André Nicolas Jul 26 '14 at 00:31
  • Thanks.. But my questions is.. how is it same as the summation value shown above? – Minu Jul 26 '14 at 00:39
  • I have written out the term by term integration, so that you can compare. You had everything absolutely right, except that you were expanding about $x=a$, when $a=0$ was intended, though not explicitly specified. In addition, you should have at least mentioned the general derivative. – André Nicolas Jul 26 '14 at 00:42
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    You are welcome. The question you asked about the $\frac{1}{1+t}=\cdots$ is briefly dealt with above by the mention of geometric series. You are probably familiar with $1+r+r^2+r^3+\cdot=\frac{1}{1-r}$ (sum of infinite geometric series). Put $r=-t$. – André Nicolas Jul 26 '14 at 00:49
  • Can this Taylor's series be used to approximate $ln(1+x)$ for any value of x? – Mathematics Dec 13 '15 at 20:58
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    @Mathematics: Buried in the answer is the restriction $|t|\lt 1$. The series we obtain therefore converges at least when $-1\lt x\lt 1$. It also happens to converge at $x=1$. It diverges when $x\le -1$ and also when $x\gt 1$. – André Nicolas Dec 13 '15 at 21:22
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    Hi Andre, sorry for commenting on an old post. As we know, the series $1-t+t^2-t^3+ \cdots$ doesn't converge uniformly in $|t|<1$. Then why we can integrate the series term by term? i.e. Why we can interchange the operations of integration and infinite summation? Thanks. – Sam Wong Apr 20 '21 at 04:24
  • @SamWong what makes you say that this series does not converge uniformly? Power series converge uniformly on a closed interval within their radius of convergence. – Little Narwhal Dec 31 '21 at 10:58
  • @LittleNarwhal Well, what I said is that the series doesn’t not converge uniformly \textcolor{red}{in $|t|<1$}. The author implicitly fixed $x$, so an uniform convergence on a closed subinterval works. Don’t mess with me please. – Sam Wong Dec 31 '21 at 11:26
  • @LittleNarwhal You stated a trivial result having little to do with my comment in a aggressive and arrogant tone (at lease in my feeling). Lil bro, relax. – Sam Wong Dec 31 '21 at 11:31
  • @SamWong Im not sure what you mean when you say the author implicitly fixed x and why you think my comment has little to do with yours. The way I was understanding the post, we can express 1/(1+t) as 1-t+t^2-... due to its power series expansion that converges uniformly for t in (-1,1). Hence we can integrate 1/(1+t) by integrating its power series term by term giving us the power series for ln(1+x), valid for x in (-1,1), given that we can also integrate 1/(1+t) by elementary means. Im sorry if my comment appeared arrogant it was not meant to be: I was asking a genuine question and still am. – Little Narwhal Jan 01 '22 at 15:51
  • @LittleNarwhal The series is not uniformly convergent in (-1,1). You may want to verify this by going through the definition. – Sam Wong Jan 01 '22 at 15:54
  • @LittleNarwhal Well, because the author didn’t write explicitly that ‘we fix an $x$’, I said he fixed $x$ implicitly. – Sam Wong Jan 01 '22 at 15:56
  • @SamWong ah fair enough do you mean that it's only uniformly convergent on [-r,r] for every r<1? – Little Narwhal Jan 01 '22 at 16:11

Note that $$\frac{1}{1+x}=\sum_{n \ge 0} (-1)^nx^n$$ Integrating both sides gives you \begin{align} \ln(1+x) &=\sum_{n \ge 0}\frac{(-1)^nx^{n+1}}{n+1}\\ &=x-\frac{x^2}{2}+\frac{x^3}{3}-... \end{align} Alternatively, \begin{align} &f^{(1)}(x)=(1+x)^{-1} &\implies \ f^{(1)}(0)=1\\ &f^{(2)}(x)=-(1+x)^{-2} &\implies f^{(2)}(0)=-1\\ &f^{(3)}(x)=2(1+x)^{-3} &\implies \ f^{(3)}(0)=2\\ &f^{(4)}(x)=-6(1+x)^{-4} &\implies \ f^{(4)}(0)=-6\\ \end{align} We deduce that \begin{align} f^{(n)}(0)=(-1)^{n-1}(n-1)! \end{align} Hence, \begin{align} \ln(1+x) &=\sum_{n \ge 1}\frac{f^{(n)}(0)}{n!}x^n\\ &=\sum_{n \ge 1}\frac{(-1)^{n-1}(n-1)!}{n!}x^n\\ &=\sum_{n \ge 1}\frac{(-1)^{n-1}}{n}x^n\\ &=\sum_{n \ge 0}\frac{(-1)^{n}}{n+1}x^{n+1}\\ &=x-\frac{x^2}{2}+\frac{x^3}{3}-... \end{align} Edit: To derive a closed for for the geometric series, let \begin{align} S&=1-x+x^2-x^3+...\\ xS&=x-x^2+x^3-x^4...\\ S+xS&=1\\ S&=\frac{1}{1+x}\\ \end{align} To prove in the other direction, use the binomial theorem or simply compute the series about $0$ manually.

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  • Can you tell me How you got the summation: $\frac{1}{1+x} = \Sigma_{n=0}^{\infty} (-1)^n x^n $? I understand the rest. – Minu Jul 26 '14 at 00:44
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    Start by convincing yourself that for |x|<1, 1/(1-x) = 1+x+x^2+.., and then replace x by -x. – Simon May 18 '17 at 06:01

Here is another method (not efficient) :

We use the fact that for all $x \in ]-1,1[$ , $\frac{1}{1+x}=\sum \limits_{n\ge 0} (-1)^n x^n$.

Then for all $x \in ]-1,1[$, we want to prove that : $\ln(1+x) =\sum \limits_{n\ge 0} (-1)^n \frac{x^{n+1}}{n+1}$.

Notice that for all $x \in[0,1[$, we have $\ln(1+x)= \int\limits_{0}^{x} \frac{1}{1+t}\mathrm{d}t$ and for all $x \in]-1,0]$, we have $\ln(1+x)= -\int\limits_{x}^{0} \frac{1}{1+t}\mathrm{d}t$. (Note that the function $t \mapsto \pm \frac{1}{1+t}$ is continuous on the compacts $[0,x]$ and $[x,0]$).

Now we can focus on the case where $x\in [0,1[$. The other case will be similar...

As the function $t \mapsto \frac{1}{1+t}\mathbb{1}_{[0,x]}(t)$ is positive and Lebesgue-measurable on $[0,1[$ we can write $\ln(1+x)= \int\limits_{0}^{1} \frac{1}{1+t}\mathbb{1}_{[0,x]}(t)\mathrm{d}t$.

Then $\ln(1+x)= \int\limits_{0}^{1} \sum \limits_{n\ge 0} (-1)^n t^n\mathbb{1}_{[0,x]}(t)\mathrm{d}t$ and we introduce for all $n\ge 0$ and for all $t\in [0,1[$ : $S_n(t,x)=\sum \limits_{k=0}^{n} (-1)^k t^k\mathbb{1}_{[0,x]}(t)$.

So $\ln(1+x)= \int\limits_{0}^{1} \sum \limits_{n\ge 0} (-1)^n t^n\mathbb{1}_{[0,x]}(t)\mathrm{d}t =\int\limits_{0}^{1} \lim\limits_{n\to+\infty} S_n(t,x)\mathrm{d}t$.

Then for all $n\ge 0$, the sequence of partial sums $S_n$ is Lebesgue-measurable on $[0,1[$ and for each $t\in[0,1[$ point-wise convergent to $S =\sum \limits_{n\ge 0} (-1)^n t^n\mathbb{1}_{[0,x]}(t)=\frac{1}{1+t}\mathbb{1}_{[0,x]}(t)$.

Moreover for all $n\ge 0$ and $t\in [0,1[$, we have $\vert S_n(t,x)\vert \le \sum \limits_{k= 0}^{n}t^k\mathbb{1}_{[0,x]}(t) \le \lim\limits_{n\to +\infty}\sum \limits_{k= 0}^{n}t^k\mathbb{1}_{[0,x]}(t)$. Because for all $k\ge 0$, the functions $t\mapsto t^{k}\mathbb{1}_{[0,x]}$ form a positive sequence of functions on $[0,1[$. That's why $\vert S_n(t,x)\vert \le \frac{1}{1-t}\mathbb{1}_{[0,x]}(t)$.

The function $t\mapsto \frac{1}{1-t}\mathbb{1}_{[0,x]}(t)$ is positive and Lebesgue-measurable and even Lebesgue-integrable on $[0,1[$ because $\int \limits_{0}^{1} =\frac{1}{1-t}\mathbb{1}_{[0,x]}(t)\mathrm{dt} = \int \limits_{0}^{x}\frac{1}{1-t}\mathrm{d}t = \ln(1-x)<+\infty$

Then using the dominated convergence theorem we can write that :

$\ln(1+x)=\int\limits_{0}^{1} \lim\limits_{n\to+\infty} S_n(t,x)\mathrm{d}t = \lim\limits_{n\to+\infty} \int \limits_{0}^{1} S_n(t,x)\mathrm{d}t = \lim\limits_{n\to+\infty} \int\limits_{0}^{1}\sum \limits_{k=0}^{n} (-1)^k t^k\mathbb{1}_{[0,x]}(t)\mathrm{d}t$

$= \lim\limits_{n\to+\infty} \sum \limits_{k=0}^{n} (-1)^k \int\limits_{0}^{1} t^k\mathbb{1}_{[0,x]}(t)\mathrm{d}t = \lim\limits_{n\to+\infty} \sum \limits_{k=0}^{n} (-1)^k \int\limits_{0}^{x} t^k\mathrm{d}t = \lim\limits_{n\to+\infty} \sum \limits_{k=0}^{n} (-1)^k \frac{x^{k+1}}{k+1} = \sum \limits_{k=0}^{+\infty} (-1)^k \frac{x^{k+1}}{k+1}$.

With the same reasoning we can deduce the same result for $x\in ]-1,0]$.

Finally for all $x\in ]-1,1[$, we have $\ln(1+x) =\sum \limits_{n\ge 0} (-1)^n \frac{x^{n+1}}{n+1}$.

NB : With properties on power series it takes four lines...

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Start with the Taylor series of the derivative $$\frac 1{1+x}=\frac 1{(1+a)+(x-a)}=\frac 1{1+a}\,\,\frac 1{ 1+\frac {x-a}{1+a}}$$ Let $t=\frac {x-a}{1+a}$ $$\frac 1{1+t}=\sum_{n=0}^\infty (-1)^n t^n$$ So $$\frac 1{1+x}=\frac 1{1+a}\,\,\sum_{n=0}^\infty (-1)^n \left(\frac{x-a}{a+1}\right)^n$$ Now, integrate $$\log(1+x)=\frac 1{1+a}\,\,\sum_{n=0}^\infty (-1)^n \int\left(\frac{x-a}{a+1}\right)^n \,dx=\frac 1{1+a}\,\,\sum_{n=0}^\infty (-1)^n \frac{1+a}{n+1} \left(\frac{x-a}{a+1}\right)^{n+1}+C$$ Simplify $$\log(1+x)=\sum_{n=0}^\infty (-1)^n \frac{1}{n+1} \left(\frac{x-a}{a+1}\right)^{n+1}+C$$ and using $x=a$, then $C=\log(1+a)$

Claude Leibovici
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