Can we calculate $2^k$ using this easy Taylor series? Trying to calculate $2^k$ by hand for $k\in[0,1]$, it's tempting to use the Taylor expansion of $x^k$ around $x=1$, to get:
$$2^k = 1^k + \frac{k (1)^{k-1}}{1!} + \frac{k(k-1) (1)^{k-2}}{2!} + \ldots =1+k +\frac{k(k-1)}{2!}+\ldots =\sum_{n=0}^\infty\binom{k}{n}$$
Unfortunately, $2$ lies exactly on the the radius of convergence $r = 1$, so in theory this may not converge.


Can we prove this converges to the correct value for all $k$? Numerically it does seem to.
What can be said about the rate of convergence? It seems quite slow. Can we bound the convergence?
Regarding convergence of the series, the $n$th term is

$$a_n = \frac{k(k-1) \ldots (k - n +1)}{n!}.$$

We have 

$$\frac{a_{n+1}}{a_n} = \frac{k-n}{n+1} = - \frac{n-k}{n+1} &lt; 0,$$

and the series is alternating for $n &gt; k.$

Note that

$$\frac{|a_{n}|}{|a_{n+1}|} = \frac{n+1}{n-k} = \frac{1+1/n}{1-k/n} = 1 + \frac{1+k}{n} +O\left(\frac1{n^2}\right),$$

and

$$\lim_{n \to \infty} \left(n \frac{|a_n|}{|a_{n+1}|}- (n+1)\right) = k &gt; 0.$$

There exists $N \in \mathbb{N}$ such that for $n &gt; N$

$$n \frac{|a_n|}{|a_{n+1}|}- (n+1) &gt; \frac{k}{2} \\ \implies |a_{n+1}| &lt; \frac{2}{k}\left(n|a_n| - (n+1)|a_{n+1}|\right).$$

Thus for all $m &gt; N$, the RHS forms a telescoping sum and 

$$\sum_{n = N}^m |a_{n+1}| &lt; \frac{2}{k}\left(N|a_N| - (m+1)|a_{m+1}|\right) &lt; \frac{2}{k}N|a_N|.$$

The series $\sum|a_n|$ is positive and bounded, and, hence, convergent. 

Therefore, the series $\sum_{n=0}^\infty\binom{k}{n}$ is absolutely convergent for $k &gt; 0$.

As an alternating series an error bound is

$$\left|\sum_{n=m+1}^\infty\binom{k}{n}\right| \leqslant \left|\binom{k}{m+1}\right|.$$
Since $\left( \begin{array}{c} k \\ n \end{array}\right)=0$ for $n &gt; k$ the series on the left side is always a finite sum. So to compute $2^k$ exact we need $k+1$ terms of this sum. So you can not speak of some sort of convergence.