<s>[INST] Taylor's Remainder $x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x) Prove that 
$\displaystyle x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x) <x-\frac{x^2}{2}+\frac{x^3}{3}$
My attempt: I can prove this by taking one side at a time and assuming $\displaystyle f(x) = \log (1+x)-x+\frac{x^2}{2}-\frac{x^3}{3}$ 
and then proving that it is a decreasing function and then the same for the other side.
But I am looking for a better solution using Lagrange's Mean Value Theorem or using Taylor's remainder.
 [/INST] It is much easier to solve this problem via integration.

If $t > 0$ then we can see that $$1 < 1 + t^{3}$$ and on dividing by $(1 + t) > 0$ we can see that $$\frac{1}{1 + t} < 1 - t + t^{2}$$ and integrating this equation in the interval $[0, x]$ (and noting that $x > 0$) we get $$\log(1 + x) < x - \frac{x^{2}}{2} + \frac{x^{3}}{3}\tag{1}$$ Further note that $$\frac{1 - t^{2}}{1 + t} = 1 - t$$ so that $$\frac{1}{1 + t} = 1 - t + \frac{t^{2}}{1 + t}$$ and integrating this equation on interval $[0, x]$ we get $$\log(1 + x) = x - \frac{x^{2}}{2} + \int_{0}^{x}\frac{t^{2}}{1 + t}\,dt\tag{2}$$ and clearly we can see that for $0 < t < x$ we have $$\frac{t^{2}}{1 + t} > \frac{t^{2}}{1 + x}$$ and hence $$\int_{0}^{x}\frac{t^{2}}{1 + t}\,dt > \int_{0}^{x}\frac{t^{2}}{1 + x}\,dt = \frac{x^{3}}{3(1 + x)}$$ and then from equation $(2)$ we get $$\log(1 + x) > x - \frac{x^{2}}{2} + \frac{x^{3}}{3(1 + x)}\tag{3}$$ Combining equations $(1)$ and $(3)$ we get the desired result.
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