Spivak, Ch. 20, Prob 16: Prove if $-1The following is a problem from Chapter 20 of Spivak's Calculus



*Prove that if $-1<x\leq 0$, then the remainder term $R_{n,0}$ for $\log{(1+x)}$ satisfies

$$|R_{n,0}|\leq \frac{|x|^{n+1}}{(1+x)(n+1)}\tag{1}$$

The solution manual says

For $-1<x\leq t\leq 0$ we have
$$0<1+x\leq 1+t\leq 1$$
$$0\leq \frac{1}{1+t}\leq \frac{1}{1+x}$$
So
$$\left | \int_0^x \frac{t^n}{1+t}dt \right | \leq \int_x^0
 \frac{|t|^n}{1+x}dt\leq \frac{|x|^{n+1}}{(1+x)(n+1)}$$

Let me try to understand this proof in more steps.
By Taylor's theorem we have
$$\frac{1}{1+x}=\sum\limits_{i=0}^{n-1} (-1)^i x^i+(-1)^n\frac{x^n}{1+x}$$
And if we integrate this expression then
$$\int_0^x \frac{1}{1+t}dt=\log{(1+x)}$$
$$=\int_0^x\left ( \sum\limits_{i=0}^{n-1} (-1)^i t^i \right )dt+\int_0^x (-1)^n\frac{t^n}{1+t} dt$$
$$=\sum\limits_{i=1}^n (-1)^i \frac{x^i}{i}+\int_0^x (-1)^n\frac{t^n}{1+t} dt$$
and thus we have the integral form of the remainder
$$R_{n,0,\log{(1+x)}}(x)=\int_0^x(-1)^n\frac{t^n}{1+t} dt$$
Note, however, that by assumption $-1<x\leq 0$.
The values of $t$ in the expression above are such that
$$0<x\leq t\leq 0$$
$$0<1+x\leq 1+t\leq 1$$
$$0<\frac{1}{1+t}<\frac{1}{1+x}$$
Thus
$$|R_{n,0,\log{(1+x)}}(x)|=\left |\int_0^x(-1)^n\frac{t^n}{1+t} dt\right |$$
$$=\left |\int_0^x \frac{t^n}{1+t} dt\right |$$
$$=\left |-\int_x^0 \frac{t^n}{1+t} dt\right |$$
$$=\left |\int_x^0 \frac{t^n}{1+t} dt\right |$$
$$\leq\left |\int_x^0 \frac{t^n}{1+x} dt\right |$$
$$=\left | \frac{-x^{n+1}}{(n+1)(1+x)} \right |$$
$$=\frac{|x|^{n+1}}{(n+1)(1+x)} $$
I find steps involving absolute values quite tricky, and am wondering if each of the steps above is correct?
In addition, is it possible to prove the result $(1)$ using the Lagrange form of the remainder?
We have
$$R_{n,0,\log{(1+x)}}(x)=\frac{(-1)^n x^{n+1}}{(1+t)^{n+1}(n+1}, t\in (0,x)$$
And as before, by assumption, $$-1<x<t<0 \implies 0<1+x<1+t<1$$
Thus
$$|R_{n,0,\log{(1+x)}}(x)|=\frac{|x^{n+1}|}{(1+t)^{n+1}(n+1)}<\frac{|x^{n+1}|}{(1+x)^{n+1}(n+1)}$$
which isn't quite what we want.
EDIT:
The following
$$\frac{1}{1+x}=\sum\limits_{i=0}^{n-1} (-1)^i x^i+(-1)^n\frac{x^n}{1+x}$$
is not due to Taylor's Theorem. It is due simply to the division of $1$ by $1+x$, which can be done using long division by hand for example.
 A: From my perspective there are two minor issues.

*

*You suggest that Taylor's Theorem implies
$$\frac{1}{1+x}=\sum_{i=0}^{n-1}(-x)^i+\frac{(-x)^n}{1+x},$$
where $R_{n-1}(x)=\frac{(-x)^i}{1+x}$ is the Lagrange form of the remainder.
Yes, the equality holds, and $R_{n-1}(x)$ is the Lagrange form of the remainder for the $(n-1)$-th order Taylor polynomial for $(1+x)^{-1}$ at $a=0$ (using the notation on Wikipedia). But Taylor's Theorem only ensures the existence of some $\xi\in[x,0]$ such that $R_{n-1}(x)=\frac{(-1)^{n}}{(1+\xi)^{n+1}}x^{n}$. The fact that $\xi=(1+x)^{1/(n+1)}-1\in(x,0)$ is revealed by the geometric series, credit where credit is due.

*There's a problem (only for odd $n$, but let's not get into this) with the step
\begin{aligned}
\Bigg|\int_x^0\frac{t^n}{1+t}\mathrm dt\Bigg|
\le\Bigg|\int_x^0\frac{t^n}{1+x}\mathrm dt\Bigg|,
\end{aligned}
which is exactly the (first) inequality. Why? Because you use $t\ge x$, so $1+\ge 1+x$, so $\frac{1}{1+t}\le\frac{1}{1+x}$, so $\frac{c}{1+t}\le$... Wait, if $c$ is negative and $a<b$, then $ac>bc$, right? There's the problem, we don't know the sign of $t^n$, it depends on the parity of $n$ (okay, we got into it). Instead of going through the case distinction to derive upper and lower bounds for the integral (which is possible, give it a try, if you want), we can use a nice property of the integral (and the absolute), namely
$$\Bigg|\int_x^0\frac{t^n}{1+t}\mathrm dt\Bigg|
\le\int_x^0\Bigg|\frac{t^n}{1+t}\Bigg|\mathrm dt
=\int_x^0\frac{|t^n|}{|1+x|}\mathrm dt
=\int_x^0\frac{|t|^n}{1+x}\mathrm dt,$$
where we used in the last step that $1+x>0$. On the other hand, we have $t<0$, which gives $|t|=-t$ (since $|t|=t$ for $t\ge 0$ and $|t|=-t$ for $t\le 0$ by definition).
So, now the integral yields
$$\Bigg|\int_x^0\frac{t^n}{1+t}\mathrm dt\Bigg|
=\int_x^0\frac{(-t)^n}{1+x}\mathrm dt
=\frac{0-(-x)^{n+1}}{(n+1)(1+x)}
=\frac{(-x)^{n+1}}{(n+1)(1+x)}.$$
Of course, if we want to, we can use $-x=|x|$ again.

After this thorough discussion, we almost know the answer to the second question: well, kind of. Taylor's Theorem only gives you what you stated (with an extra closing bracket for $n+1$). But, if you use that the derivative of $\ln(1+x)$ is $(1+x)^{-1}$ and the geometric series, well, then, as we know, the correct choice of $t$ (as given by the mean value theorem) is $(1+x)^{1/(n+1)}-1$.
Just to be very, very clear here, for whoever this reads: For no form of the remainder in Taylor's Theorem can we ever choose the "$\xi$ between $x$ and $a$", never. Never. (Of course, unless the mean is attained at several values in the interval, then we can choose out of these, and what a fine choice this is).
