The following question is from Stephen Abbott's "Understanding Analysis."
Question: Explain how Lagrange’s Remainder Theorem can be modified to prove $$ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4} \cdots = \ln2. $$
A question close to mine was asked here but does not address my concern.
The usual procedure used to solve such problems is to first show that the power series $f(x) = \sum_{k=1}^\infty \frac{(-1)^{k-1}x^{k}}{k}$ can be derived from that of $1/(1+x)$ so that the region of convergence for $f(x)$ would be $(-1,1)$, then to show that $f(x)$ converges at $x=1$ (using alternating series test in this case). Abel's theorem would imply that $f(x)$ converges uniformly on $[0,1]$ and continuous limit theorem would imply that $f(x)$ is continuous at $x=1$. Finally, as $f(x) = \ln(1+x)$ on $(-1,1)$, we can conclude that $\ln(2) = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4} \cdots$
The question however asks us to modify Lagrange’s Remainder Theorem. Following are two statements, the first of which is Lagrange’s Remainder Theorem as given in the text, and the second is the one that I have modified. Please let me know if second statement is correct and if not, please give me a hint as to where I am going wrong.
Lagrange’s Remainder Theorem (as given in the text): Let $f$ be differentiable $N + 1$ times on $(−R,R)$. Define $a_n = f^{(n)}(0)/n!$ for $n = 0,1,\cdots,N$, and let $$ S_N(x) = a_0 +a_1x+a_2x^2 +\cdots+a_Nx^N. $$ Given $x\neq 0$ in $(−R,R)$, there exists a point $c$ satisfying $|c| < |x|$ where the error function $E_N(x) = f(x) − S_N(x)$ satisfies $$ E_N(x) = \frac{f^{(N+1)}(c)x^{N+1}}{(N + 1)!}. $$
Modified Lagrange’s Remainder Theorem: Let $f$ be differentiable $N + 1$ times on $(−R,R)$, be continuous at $x=R$, and let the Taylor series of $f$ be convergent at $x=R$ ($R>0$). Define $a_n = f^{(n)}(0)/n!$ for $n = 0,1,\cdots,N$, and let $$S_N(x) = a_0 +a_1x+a_2x^2 +\cdots+a_Nx^N.$$ There exists a point $c$ satisfying $|c| < R$ where the error function $E_N(x) = f(x) − S_N(x)$ satisfies $$ E_N(R) = \frac{f^{(N+1)}(c)R^{N+1}}{(N + 1)!}. $$
The proof basically runs along the same lines as the first approach that I have outlined above which uses Abel's Theorem, along with the standard proof for Lagrange’s Remainder Theorem.