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Show that for all $x>0$ we have $\ln(1+x)>x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$.

I know it has to do with taylor expansion but somehow I cannot prove it rigorously.

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Take the derivative of the LHS minus the RHS to get

$$\frac{1}{1+x} - 1 + x - x^2 + x^3 = \frac{x^4}{x+1}$$ which is clearly nonnegative.

So the difference is monotone increasing. Since both sides are equal at $0$, the LHS is always $\ge$ the RHS.

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Hint: the two sides are equal at $x=0$. What about the derivatives?

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Another way of showing that, having in mind that $\ln(x)$ is differentiable infinite many times on $(0,\infty)$, is looking at the remainder. First - $$\ln(1+x)=\sum_{k=1}^n (-1)^{k+1} \frac{x^k}{k!} +R(c,n)$$ Where $R(c,n)=\frac{[\ln(1+x)]^{(n+1)}\big|_{x=c}}{(n+1)!}\cdot x^{n+1}$ for $c \in [0,x]$

When plugging $n=4$ we get that $R(n,c)>0$, as desired.

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