Show that $\ln(1+x)\leq x-{1\over 2}x^2+{1\over 3}x^3$. Prove that for $x\in (0,\infty)$, $\ln(1+x)\leq x-{1\over 2}x^2+{1\over 3}x^3$.
I'm a little bit stuck, but I think I have the right idea. Any hints or solutions are greatly appreciated. Here is what I have.
If we assume that $\ln(1+x)>x-{1\over 2}x^2+{1\over 3}x^3$ and let $f(x)=\ln(1+x)-x+{1\over 2}x^2-{1\over 3}x^3>0$, then $f'(x)={1\over{1+x}}-1+x-x^2={-x^3\over 1+x}<0$ for all $x>0$. That is, $f'(x)<0$ implies that $f(x)$ decreasing which is a contradiction.
Does this make sense?
 A: You could convert yours into a direct proof easily. As you have, let
$$f(x) = \ln(1+x) - x + \dfrac{x^2}2 - \dfrac{x^3}3$$
We then have
$$f'(x) = \dfrac1{1+x} - 1 + x - x^2 = \dfrac{-x^3}{1+x} < 0 \text{ since }x \in (0,\infty)$$
Hence, $f(x)$ is a decreasing function, which means
$$f(x) < f(0) \implies \ln(1+x) - x + \dfrac{x^2}2 - \dfrac{x^3}3 < 0 \implies \ln(1+x) < x - \dfrac{x^2}2 + \dfrac{x^3}3$$
A: Start with
$\ln (1+x)
=\int_1^{1+x} \frac{dt}{t}
=\int_0^{x} \frac{dt}{1+t}
$
for $x \ge 0$.
The finite series expansion
for $\frac1{1+t}$
is, for any $n$,
$\frac1{1+t}
=\sum_{k=0}^{n-1} (-1)^k t^k + \frac{(-1)^n t^n}{1+t}
$.
Substituting this,
$\begin{array}\\
\ln (1+x)
&=\int_1^{1+x} \frac{dt}{t}\\
&=\int_0^{x} \frac{dt}{1+t}\\
&=\int_0^{x} \left(\sum_{k=0}^{n-1} (-1)^k t^k + \frac{(-1)^n t^n}{1+t}\right)dt\\
&=\sum_{k=0}^{n-1} \frac{(-1)^k x^{k+1}}{k+1} + \int_0^{x}\frac{(-1)^n t^n}{1+t}dt\\
&=\sum_{k=1}^{n} \frac{(-1)^{k-1} x^{k}}{k} + R_n(x)\\
\end{array}
$
where
$R_n(x)
=\int_0^{x}\frac{(-1)^n t^n}{1+t}dt
$.
Now, watch carefully - 
at no time do the fingers leave the hands.
If $n$ is even,
$R_n(x)\ge 0$
and if $n$ is odd,
$R_n(x)\le 0$,
because the sign of the integrand
depends only on $(-1)^n$.
Therefore
$R_{2n+1}
\le 0
\le R_{2n}
$,
so that
$$\ln (1+x)-\sum_{k=1}^{2n+1} \frac{(-1)^{k-1} x^{k}}{k}
\le 0
\le \ln (1+x)-\sum_{k=1}^{2n} \frac{(-1)^{k-1} x^{k}}{k}
$$
or
$$\sum_{k=1}^{2n} \frac{(-1)^{k-1} x^{k}}{k}
\le \ln(1+x)
\le \sum_{k=1}^{2n+1} \frac{(-1)^{k-1} x^{k}}{k}
.
$$
Setting $n=1$,
$$x-\frac{x^2}{2}
\le \ln(1+x)
\le x-\frac{x^2}{2}+\frac{x^3}{3}
.$$
Note that this shows that the series
$\sum_{k=1}^{n} \frac{(-1)^{k-1} x^{k}}{k}
$
is an enveloping series
for $\ln(1+x)$,
because the partial sums
are alternatingly
above and below
$\ln(1+x)$,
and this holds even when the
series does not converge.
As in many of my answers,
this one is not original.
It is taken from
chapter 14
of Heinrich Dorrie's
most excellent book
"100 Great Problems of
Elementary Mathematics."
This book has similarly understandable
derivations of the series for
$e^x$, $\sin(x), \cos(x),
\arctan(x), \tan(x),
$ and
$\sec(x)$,
as well as the solutions
to many other well-known problems.
This book is available for
about \$15 US from Dover
and
\$10 US from Google books.
I have both versions,
and highly recommend it.
Don't just sit there -
get your own copy!
