how to prove that $\ln(1+x)< x$

Question

I want to prove that: $\ln(x+1)< x$.

My idea is to define: $f(x) = \ln(x+1) - x$, so:

$f'(x) = \dfrac1{1+x} - 1 = \dfrac{-x}{1+x} < 0, \text{ for }x >0$.

Which leads to $f(x)<f(0)$, so $\ln (x+1)-x<0$.

Is that a valid proof? Any other ideas?

Thanks.

Answer 1

I think your approach is correct but you need to add some more details. Based on your approach let $f(x) = \log(1 + x) - x$ so that $f(0) = 0$. Clearly $$f'(x) = -\frac{x}{1 + x}$$ and hence $f'(x) > 0$ if $-1 < x < 0$ and $f'(x) < 0$ if $x > 0$. It follows that that $f(x)$ in increasing in $(-1, 0]$ and decreasing in $[0, \infty)$. Thus we have $f(x) < f(0)$ if $-1 < x < 0$ and $f(x) < f(0)$ if $x > 0$. It thus follows that $f(x) \leq f(0) = 0$ for all $x > -1$ and there is equality only when $x = 0$. So we can write $$\log(1 + x) \leq x$$ for all $x > -1$ and there is equality only when $x = 0$.

Note: We have considered $x > -1$ because $\log(1 + x)$ is not defined if $x \leq -1$.

Answer 2

I think it does work for $x>0$, but also it works for $x > -1$. Notice that $f'$ tells you that $f$ has an absolute maximum at $x=0$ and so $f(x) \leq f(0)=0$ for all $x>-1$.

Answer 3

Assuming we are working on $x>0 $ we have $$\log\left(x+1\right)<x\Leftrightarrow x+1<e^{x}\Leftrightarrow x+1<\sum_{k\geq0}\frac{x^{k}}{k!}\Leftrightarrow1<\sum_{k\geq1}\frac{x^{k-1}}{k!} $$ and this is true because $$\sum_{k\geq1}\frac{x^{k-1}}{k!}=1+\sum_{k\geq2}\frac{x^{k-1}}{k!}.$$ Note that for $x=0$ the inequality fails.

Answer 4

Another proof is based on the fact that $e^x$ is a convex function and $x+1$ is tangent to $e^x$ at $0$. That is,

$$x+1< e^x, \text{ if } x\not =0, \text { and } x+1=e^x \text{ if } x=0.$$

Taking the natural logarithm of both sides we get that

$$\ln(x+1)< x \text { if } x\not =0.$$

Answer 5

$$e^x = 1 + x + \frac{x^2}{2}+ \frac{x^3}{6}+ \cdots > 1+x$$

if $x>0$. Then, taking the logarithm, which is an increasing function, we get $x > \ln(1+x)$.