Is the sum of the absolute value of the even terms in a Maclaurin Series greater than the sum of the odd terms? 
(Please see heading) - also my context is: show that $$x\ge0$$
  $$x-\frac{x^2}{2}\le \ln(1+x)\le x-\frac{x^2}{2}+\frac{x^3}{3}$$
  It's obvious that the first term is smaller than the last term, but how do I show what the size of the second term is - I am aware of the Maclaurin series and already have seen its expansion for $$\ln(1+x)$$

 A: The answer to the question in the heading is in general no.  For this particular series, the answer is yes.  The simplest way to see this is
to note that for $t \ge 0$,
$$1-t^2 \le 1 \le 1 + t^3 $$
so that
$$ 1 - t \le \dfrac{1}{1+t} \le 1 - t + t^2$$
Integrate for $t$ from $0$ to $x$, and you find that for $x \ge 0$,
$$ x - \dfrac{x^2}{2} \le \ln(1+x) \le x - \dfrac{x^2}{2} + \dfrac{x^3}{3}$$
And more generally, for any positive integer $k$, if $x \ge 0$ 
$$ \sum_{j=1}^{2k} (-1)^{j-1} \dfrac{x^j}{j} \le \ln(1+x) \le
\sum_{j=1}^{2k+1} (-1)^{j-1}  \dfrac{x^j}{j} $$
A: Using integration by parts, Taylor's theorem with the integral form of the remainder is
$$ f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k + R_n(x), $$
where
$$ R_n(x) = \int_a^x \frac{(x-t)^n}{n!} f^{(n+1)}(t) \, dt. $$
What this problem is getting at is that the derivatives of $\log{(1+x)}$, which I'm sure you know are
$$ \frac{1}{1+x} , \frac{-1}{(1+x)^2}, \frac{2}{(1+x)^3}, \dotsc, $$
alternate in sign. Thus the expression in the integral is positive for even $n$ and negative for odd $n$. Therefore
$$ f(x) - \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k = R_n(x) $$
is positive if $n$ is even and negative if $n$ is odd, since the integral of a positive function is positive (and the same with positive replaced by negative). Then the inequalities fall straight out, and indeed, generalisations thereof to more terms in the expansion.
