Using only Lagrange’s Remainder Theorem (and no references to Abel’s Theorem) prove $1 − 1/2 +1/3 − 1/4 +1/5 − 1/6 + ··· = \ln(2)$.
As I understand, the Lagrange Theorem states, that if the remainder $f^{(n+1)}(c)x^{n+1}/(n+1)!$ converge than the series converge uniformly.
How can I use it here? What is $x$ in this case? I have that $f(x) = \ln(x)$ and at $x=2$ it converges, is it right?
Look at the Taylor polynomial of $f(x) = \ln x$ at $a = 1$ with the Lagrange form of the remainder. Since $f$ is infinitely differentiable on $(0,\infty)$, for any $n \ge 1$ and for any $x >0$ you have
$$f(x) = \sum_{k=0}^n \frac{f^{(k)}(1)}{k!}(x-1)^k + \frac{f^{n+1}(\xi_n)}{(n+1)!} (x-1)^{n+1}$$ for some $\xi_n$in between $1$ and $x$. Since $f(1) = 0$ and $$f^{(k)}(x) = (-1)^{k-1} \frac{(k-1)!}{x^k}$$ for all $x > 0$ you can evaluate the above expression becomes $$\ln x = \sum_{k=1}^n\frac{(-1)^k}{k} (x-1)^k + \frac{(-1)^n}{(n+1) \xi_n^{n+1}}(x-1)^{n+1}.$$
If $x=2$ then $\xi_n \ge 1$ so that $$ \left| \ln x - \sum_{k=1}^n\frac{(-1)^k}{k} (x-1)^k \right| \le \frac{1}{n+1}.$$ Now take $n \to \infty$.