Exercise 6.6.4

Let $E_N(x)=\log <!--\; FUNCTION\; APPLICATION\; -->(1+x)-{\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }\frac{{(-1)}^n{x}^n}{n}$. The proof presented of Lagrange’s Remainder Theorem assumed that $x>0$ to simplify notation, and found a value of $c\in (0,x)$, with the proof for $x<0$ being implicit. But we can just ignore $x<0$ to state that given $x\in (0,R)$ there exists $0<c<x$ satisfying

Now, $\left|f^{(N+1)}(c)\right|=N!{(1+c)}^{-(N+1)}$, so for $c<x=1$ we have

which converges to 0 as $N\to \infty$. Hence the Taylor series is equal to $\log <!--\; FUNCTION\; APPLICATION\; -->(1+x)$ over at least $(0,1)$, and we can extend this equality to $(0,1]$ since both $\log <!--\; FUNCTION\; APPLICATION\; -->(1+x)$ and the Taylor series are continuous at 1 (we established the latter in Exercise 6.5.1). Plugging $x=1$ into both equations leaves us with the desired equality.