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.
 A: 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.
A: This is shown in
section 14 of
Heinrich Dorrie's
"100 Great Problems of
Elementary Mathematics",
which I highly recommend.
The proof,
for general $n$,
goes like this:
Let
$f(t) = \dfrac1{1+t}$.
Then,
as can be shown by induction,
$$f(t)
=\sum_{k=0}^{n-1} (-1)^{k}t^k+(-1)^nt^nf(t)
$$
Integrating from
$0$ to $x$,
\begin{align}\\
\ln(1+x)
&=\int_0^x \frac{dt}{1+t}\\
&=\int_0^x \left(\sum_{k=0}^{n-1} (-1)^{k}t^k+(-1)^nt^nf(t)\right)dt\\
&=\sum_{k=0}^{n-1} \int_0^x(-1)^{k}t^kdt+\int_0^x(-1)^nt^nf(t)dt\\
&=\sum_{k=0}^{n-1} (-1)^{k}\frac{x^{k+1}}{k+1}+(-1)^n\int_0^x\frac{t^ndt}{1+t}\\
&=\sum_{k=0}^{n-1} (-1)^{k}\frac{x^{k+1}}{k+1}+(-1)^nr_n(t)\\
\end{align}
where
$$r_n(t)
=\int_0^x\frac{t^ndt}{1+t}
$$
Since
$1 \le 1+t
\le 1+x
$ for
$0 \le t \le x$,
$$t^n 
\ge \frac{t^n}{1+t}
\ge \frac{t^n}{1+x}
$$
so that,
since
$$\int_0^x t^n dt
=\frac{x^{n+1}}{n+1}
$$
$$\frac{x^{n+1}}{n+1}
\ge r_n(x)
\ge \frac{x^{n+1}}{(n+1)(1+x)}
$$
Therefore
$$r_n(x)
=\frac{x^{n+1}}{(n+1)(1+v)}
$$
where
$0 < v < x
$.
Therefore
$$\ln(1+x)
=\sum_{k=0}^{n-1} (-1)^{k}\frac{x^{k+1}}{k+1}
+(-1)^n\frac{x^{n+1}}{(n+1)(1+v)}
$$
where
$0 <v < x$.
A: For the inequality on the right, note that the Lagrange form of the remainder gives
$$\log (1+x) = x- \frac{x^2}{2} +\frac{x^3}{3} -\frac{1}{4(1+c)^4},$$
where $0<c<x.$ Since the last term in negative, we get the strict inequality. In fact we get the strict inequality for $-1<x<0$ as well, by the same argument.
The inequality on the left seems easier for $-1<x<0$ to me. Here we get
$$\log (1+x) = x-x^2/2 + x^3/3 + \cdots = -(|x|+|x|^2/2 + |x|^3/3+ |x|^4/4\cdots )$$ $$> -(|x|+|x|^2/2 + (|x|^3/3)(1+|x| + |x|^2 + \cdots))$$ $$ = -(|x|+|x|^2/2 + (|x|^3/3)/(1-|x|) = x-x^2/2 + (x^3/3)/(1+x).$$
So there is good motivation for looking at $x-x^2/2 + (x^3/3)/(1+x)$ as below $\log (1+x)$ for all $x\in (-1,\infty), x\ne 0.$ At this point I would do exactly as you did in showing this is true for $x>0.$
