Accurately evaluating the hypergeometric function

Before people get any ideas: although we have the identity:

$${}_2 F_1\left({{1,1}\atop{m}}\mid -1\right)=\frac{m-1}{2}\Phi\left(\frac12,1,m-1\right)$$

where $\Phi(z,s,a)$ is the Lerch transcendent; or, in Mathematica notation:

Hypergeometric2F1[1, 1, m, -1] == (m - 1) HurwitzLerchPhi[1/2, 1, m - 1]/2

the computation becomes even more unstable with that replacement (I'm not really sure why); don't use it.

In any event, we have the relations

$$\begin{align*} {}_2 F_1\left({{1,1}\atop{m}}\mid -1\right)&=2^{m-2}(m-1)\sum_{k=m-1}^\infty \frac1{k 2^k}\\ &=2^{m-2}(m-1)\left(\log\,2-\sum_{k=1}^{m-2}\frac1{k 2^k}\right) \end{align*}$$

where we see why N[] might have a spot of trouble with evaluating the exact expression produced by Hypergeometric2F1[]: $\sum\limits_{k=1}^{m-2}\frac1{k 2^k}\approx \log\,2$, with the difference getting smaller as $m\to\infty$, and we thus see a fair amount of catastrophic cancellation during numerical evaluation. In particular, for $m=200$, $\sum\limits_{k=1}^{m-2}\frac1{k 2^k}$ and $\log\,2$ agree to $61$ (!) decimal places.

Fortunately for us, Hypergeometric2F1[] can cope nicely with inexact arguments:

Hypergeometric2F1[1, 1, N[200, 20], -1]
   0.9950490265763910737

In short: just supply inexact numbers to Hypergeometric2F1[] at the outset.