The proof of $\log(1+x) < x.$ I want to prove $\log(1+x) < x.$
Of course, I can prove this by letting $f(x):= x - \log(1+x),$ and calculating $f'(x)$, ... .
But I wonder if I can prove this using Maclaurin's expansion.
$\log(1+x)=x-\dfrac{1}{2}x^2+\dfrac{1}{3} x^3-\cdots.$
The right-hand side includes the term "$x$" so it seems that I can use this equality in order to prove $\log(1+x) < x$, but I cannot.
I would like you to give me some ideas.
 A: Since you talk about series expansions, one could exponentiate the inequality (the exponential function is monotonically increasing, so this is equivalent to the original inequality) to obtain $e^x > 1 + x$, which is easy to prove with the series expansion of the exponential function: we have $$e^x = 1 + x + \underbrace{x^2 + \ldots}_{> 0} > 1 + x$$ for all $x > 0$.
A: Take $0<x<1$, then if we let $a_k(x) = (-1)^{k+1}\frac{1}{k}x^k$ we see that
$$\log (1+x)-x = \sum_{k=2}^{\infty}a_k(x).$$
Now note that in the specified interval we have that $|a_k(x)|>|a_{k+1}(x)|$ and that $\mathrm{sgn}\, a_k(x) = -\mathrm{sgn}\, a_{k+1}(x)$ so that the sum is alternating. These two properties implies that $a_{2k}(x)+a_{2k+1}(x)<0$ and therefore
$$\log(1+x)-x= \sum_{k=2}^{\infty}a_k(x) = \sum_{k=1}^{\infty}\underbrace{a_{2k}(x)+a_{2k+1}(x)}_{<0}<0$$
A: For me the simplest proof is to look for the minimum value of $$f(x)=x- \log (1+x)$$
$$f'(x)=1-\frac 1{1+x} \qquad \text{and} \qquad f''(x)=\frac 1{x^2} \quad > 0~~\forall x$$  So the extremum is at $x=0$ and the second derivative test shows that this is a minimum. So, in real domain, $\forall x$
$$x \geq \log(1+x).$$
A: The definition of natural log in How can we come up with the definition of natural logarithm? allows a proof without words.

interactive graph made from Desmos
\begin{align}
\mathrm{LHS} &:= \int_1^{1+x} \frac{\mathrm{d}t}{t} = \text{area under the curve} \\
\mathrm{RHS} &= x = \text{area of the rectangle}
\end{align}
The curve $y = 1/x$ is below $y = 1$ on $x \ge 1$.
Exercise: complete this proof for the case $x \in (-1,0)$.
A: If you accept (otherwise this can easily be proved) that $\log ()$ is a concave function, then it suffices to show (cf. Jensen) that $x$ is a tangent to $\log(1+x)$. But this is obvious: $x$ and $\log(1+x)$ touch at $x=0$ and have the same slope ($1$) there.
A: A "proof without words" : $\forall x>-1, x\neq 0, \log(1+x)<x.$
