Verifying This Proof for Alternating Harmonic Series

All the matter stems from that , for real $r$, the expression $$ \left( {1 + x} \right)^{\,r} = \sum\limits_{0\, \le \,k} { \binom{r}{k} x^{\,k} } = \sum\limits_{0\, \le \,k} {{{r^{\,\underline {\,k\,} } } \over {k!}}x^{\,k} } \quad \left| {\;r \in \mathbb R} \right. $$ (where we indicate with $r^{\,\underline {\,k\,} }$ and $r^{\,\overline {\,k\,} }$, respectively, the Falling and Rising Factorial)

• converges absolutely , for whichever $r$, if $|x|<1$;
• for $x=1$ , it converges for $-1<r$;
• for $0 \le r \in \mathbb Z$ the sum is finite, and thus converges absolutely, for whichever $x$.
  re. to this article in Wikipedia.
  Note in fact that if $r$ is not a non-negative integer, the sum will contain infinitely many terms with alternated sign.

We know that we are allowed to drift inside the sum some algebraic manipulation (including taking the limit) if the sum converges absolutely, while if the sum is just convergent then the convergence might be compromised.
Therefore let's proceed cautiously

For $|x|<1$ we can write $$ \eqalign{ & {{\left( {1 + x} \right)^{\,r} - 1} \over r} = \sum\limits_{1\, \le \,k} {{{r^{\,\underline {\,k\,} } } \over {r\,k!}}x^{\,k} } = \sum\limits_{0\, \le \,k} {{{r^{\,\underline {\,k + 1\,} } } \over {r\,\left( {k + 1} \right)!}}x^{\,k + 1} } = \cr & = \sum\limits_{0\, \le \,k} {{{\left( {r - 1} \right)^{\,\underline {\,k\,} } } \over {\left( {k + 1} \right)!}}x^{\,k + 1} } = x\sum\limits_{0\, \le \,k} {{1 \over {k + 1}}{{\left( {r - 1} \right)^{\,\underline {\,k\,} } } \over {k!}}x^{\,k} } = \int_0^x {\left( {1 + t} \right)^{\,r - 1} dt} \cr} $$ and the integral indicates that, for $r$ approaching $0$, we are running over the edge $$ \int {t^{\,r - 1} dt} = {1 \over r}t^{\,r} \quad \int {t^{\, - 1} dt} = \ln (t) $$ At the same time, the integral is well defined for $x \to 1^{-}$ and for $r \to 0$.

So $$ \eqalign{ & \mathop {\lim }\limits_{r\; \to \,0} {{\left( {1 + x} \right)^{\,r} - 1} \over r} = \ln \left( {1 + x} \right) = \sum\limits_{0\, \le \,k} {{{\left( { - 1} \right)^{\,\underline {\,k\,} } } \over {\left( {k + 1} \right)!}}x^{\,k + 1} } = \cr & = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} {{1^{\,\overline {\,k\,} } } \over {\left( {k + 1} \right)!}}x^{\,k + 1} } = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} {{k!} \over {\left( {k + 1} \right)!}}x^{\,k + 1} } = \cr & = \sum\limits_{0\, \le \,k} {{{\left( { - 1} \right)^{\,k} } \over {\left( {k + 1} \right)}}x^{\,k + 1} } \cr} $$ which is the well known Mercator series, and known to be convergent for $-1<x \le 1$.
Note: above we made use of the fact that, for whichever integer $k$ and real (or complex) $s$ we have $$ \eqalign{ & \left( { - s} \right)^{\,\underline {\,k\,} } = \left( { - s} \right)\left( { - s - 1} \right) \cdots \left( { - s - \left( {k - 1} \right)} \right) = \cr & = \left( { - 1} \right)^{\,k} s\left( {s + 1} \right) \cdots \left( {s + \left( {k - 1} \right)} \right) = \left( { - 1} \right)^{\,k} s^{\,\overline {\,k\,} } \cr} $$

So we can take the limit for $x \to 1^{-}$, and obtain $$ \mathop {\lim }\limits_{x\; \to \,1^{\, - } } \ln \left( {1 + x} \right) = \ln 2 = \sum\limits_{0\, \le \,k} {{{\left( { - 1} \right)^{\,k} } \over {\left( {k + 1} \right)}}} $$

There are plenty of posts herewith dealing with this sum, but refer in particular to this post to understand how "delicate" it is: you cannot rearrange the terms (for instance).

To expand on the other answers, you are missing the argument that this also holds for $n>0$ real (else you cannot just take the real limit to $0$). This can be solved by using the binomial series.

$$2^n=1+n+\frac{n(n-1)}{2!}+\frac{n(n-1)(n-2)}{3!} +...$$ is indeed true for any $n\in\mathbb{Z}_{>0}$. Looking at the Maclaurin series of $$(1+x)^n$$ with $n>0$ given by $$1+nx+\frac{n(n-1)x^2}{2!}+\ldots$$ which is convergent in $x=1$, since $\text{Re}(n)>-1$. Now we get exactly $$2^n=1+n+\frac{n(n-1)}{2!}+\frac{n(n-1)(n-2)}{3!} +...$$ for all $n>0$. Indeed, this implies $$\frac{2^n-1}{n}=1+\frac{(n-1)}{2!}+\frac{(n-1)(n-2)}{3!}+...:=\sum_{k=1}^\infty f_n(k)$$ for $0<n<1$ (we choose this interval because then the sum will be infinite: only for $n$ integer it's finite).

Now the big problem is that we need to swap a limit and an infinite sum on the RHS. If we have this, then indeed $$\ln(2)=\lim_{n\downarrow 0}\frac{2^n-1}{n} =\lim_{n\downarrow 0}\sum_{k=1}^\infty f_n(k)=\sum_{k=1}^\infty \lim_{n\downarrow 0}f_n(k)=1-\frac{1}{2}+\frac{1}{3}+...$$ However, this cannot be solved by the dominated convergence theorem: we need to find some function $g:\mathbb{N}\to\mathbb{R}$ such that $|f_n(k)|\leq g(k)$ for all $k\in\mathbb{Z}>0$ and $\sum_{k=1}^\infty g(k)$. This means that $g(k)$ will at least be the harmonic series, which does not converge.

Assuming that $1-\frac{1}{2}+\frac{1}{3}+...$ converges, we have pointwise convergence on compact interval $[0,\frac{1}{2}]$. Dini's theorem doesn't work either, since the series is alternating, so not increasing/decreasing.

Trying to prove uniform convergence also requires estimates with absolute values, resulting in the harmonic series.

With these tools it doesn't work, so this must be such a case where something is true, but the tools break when trying them out on your specific proof. I conclude that you just got lucky somewhere along the way.

The equality$$2^n=\sum_{k=0}^n\binom nk$$holds when $n\in\mathbb{Z}_+$. It doesn't make sense to talk about$$\lim_{n\to0}\frac{2^n-1}n$$using it. Besides, how did your factorials vanished?