I am reading "Calculus 4th Edtion" by Michael Spivak.
On p.427 he wrote as follows:
From the calculations on page 413, we see that for $x \geq 0$ we have $$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots+\frac{(-1)^{n-1}x^n}{n}+\frac{(-1)^n}{n+1}t^{n+1}$$ where $$\left \vert {\frac{(-1)^n}{n+1} t^{n+1}}\right\vert \leq \frac{x^{n+1}}{n+1}$$ and there is a slightly more complicated estimate when $-1 < x < 0$ (Problem 16).
I guess $t \in (0, x)$ if $x > 0$ and $t = 0$ if $x = 0$.
Does the above equality really hold?
I guess Spivak applied Taylor's theorem(p.424, Lagrange form of the remainder) incorrectly to $\log(1 + x)$.
Actually, what Spivak writes is that$$\log(1+x)=x-\frac{x^2}2+\frac{x^3}3-\cdots+(-1)^{n-1}\frac{x^n}n+\int_0^x\frac{t^n}{t+1}\,\mathrm dt.$$But then$$x\geqslant0\implies\int_0^x\frac{t^n}{t+1}\,\mathrm dt\leqslant\int_0^xt^n\,\mathrm dt=\frac{x^{n+1}}{n+1}.$$