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For $|x|<1$, we have that

$$ \ln(1+x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \cdots $$

Is there any elementary(try not to use integration or differantiation) proof for the equality above?

Edit: I have changed my definition of elemantary.

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    $\begingroup$ @user251257 As the title suggests, "natural logarithm". $\endgroup$
    – user429582
    Commented Apr 17, 2018 at 20:00
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    $\begingroup$ there are many ways to define the natural logarithm, thus the answer depends on your definition. $\endgroup$
    – user251257
    Commented Apr 17, 2018 at 20:03
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    $\begingroup$ How do you even know the natural logarithm exists without using calculus? $\endgroup$
    – saulspatz
    Commented Apr 17, 2018 at 20:12
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    $\begingroup$ @BrightChancellor but you can't, and the proof below uses integration. You still haven't shown us that you can prove this using integration. $\endgroup$
    – amWhy
    Commented Apr 17, 2018 at 20:19
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    $\begingroup$ @BrightChancellor, if your definition of the logarithm is the inverse of the exponential, how do you define the exponential? $\endgroup$
    – Yuriy S
    Commented Apr 17, 2018 at 21:07

2 Answers 2

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Here is a sketch, which if extended will find any particular coefficient without calculus, on the assumption a polynomial expression exists

Let's start with the definitions:

  • $\exp(x)= 1+\frac{1}{1!}x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots$ and
  • for positive $x$ we have $\exp(\ln(x))=x$

Suppose $\ln(1+x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots$

Clearly $a_0=0$ since $\exp(0)=1$ and so $\ln(1+0)=0$

Then we need $1+x = 1 + \frac{1}{1!}(a_1x+a_2x^2+a_3x^3+\cdots)+\frac{1}{2!}(a_1x+a_2x^2+\cdots)^2+\frac{1}{3!}(a_1x+\cdots)^3+\cdots$

which by

  • matching coefficients of $x$ will give $a_1=1$, and
  • matching coefficients of $x^2$ will give $a_2+\frac{a_1^2}{2!}=0$ so $a_2=-\frac12$, and
  • matching coefficients of $x^3$ will give $a_3+\frac{2a_1a_2}{2!}+\frac{a_1^3}{3!}=0$ so $a_3=+\frac13$, and
  • matching coefficients of $x^4$ will give $a_4+\frac{2a_1a_3+a_2^2}{2!}+\frac{3a_1^2a_2}{3!}+\frac{a_1^4}{4!}=0$ so $a_4=-\frac14$, and
  • we can do something similar for later coefficients
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  • $\begingroup$ Although this is implicitly using a theorem that if two infinite power series converge to the same values for all $x$ in some interval, then the series must have the same coefficients - which could be difficult to prove without differentiation. $\endgroup$
    – aschepler
    Commented Apr 18, 2018 at 4:57
  • $\begingroup$ @aschepler Well, to avoid that we could forget issues of convergence and derive these coefficients for $\exp$ and $\ln$ viewed as formal series that are inverses. (Of course, that gives the coefficients, but relating the formal series to the real functions may be a challenge without using differentiation and integration.) $\endgroup$
    – ShreevatsaR
    Commented Apr 18, 2018 at 5:55
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Consider

$$ \frac{1}{1-x} = \sum_{n \geq 0 } x^n $$

Thus,

$$ \frac{1}{1+x} = \sum_{n \geq 0 } (-1)^nx^n $$

Integrating, we obtain

$$ \ln (1+x) + C = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1} }{n+1} $$

With $x=0$ , $C$ better be zero and thus

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

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  • $\begingroup$ Beautiful, thank you. But is there a calculus-free proof for your first line? Also, you used integration. $\endgroup$
    – user429582
    Commented Apr 17, 2018 at 20:01
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    $\begingroup$ @BrightChancellor: how do you define $\log$ without Calculus? $\endgroup$
    – Jack D'Aurizio
    Commented Apr 17, 2018 at 20:11
  • $\begingroup$ @JackD'Aurizio You're right, I can't. $\endgroup$
    – user429582
    Commented Apr 17, 2018 at 20:13
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    $\begingroup$ The OP said "try not to use integration or differentiation". $\endgroup$
    – Fly by Night
    Commented Apr 17, 2018 at 20:19