Prove that $ \ln[e(2/e)] $ is a fast way to calculate $ \ln2 $ Consider formula  $ (*) \ln(x) = \sum_{k=1}^{\infty} (-1)^{k-1}\cdot \frac{(x-1)^k}{k}$. If you calculate $ \ln2 $ with error less then $ \frac{1}{2} \cdot 10^{-6} $ we need more than two milion element's series $ (*) $. Prove that $ \ln2 = \ln[e(2/e)] $ is much faster.
Idea: 
I guess, that formula $ (*) $ is correct for very small interval, $x \in (-1,1) $ 
So $ \ln[e(2/e)]=\ln e+\ln(2/e) = 1 + \ln(2/e) $
And know: 
- how prove that $ \ln(2/e) \in (-1,1) $ 
- I must consider something else? Is in the idea is a trap?
 A: The transcendental constant $\ln 2$ is useful in many applications, cf. range reduction in computing $e^x$ or converting between natural logarithms and logarithms base two.
It might be more difficult to compute $e$ or its inverse "from scratch" to six decimal places than to use some other method of improving a rate of convergence.
Let's consider how close $2/e$ gets us to the expansion point around $x=1$:
$$ | 2/e - 1 | \approx 0.264 $$
Another approach would be to take advantage of square roots, since $\ln 2^\frac{1}{2} = \frac{1}{2} \ln 2$:
$$ | 2^{\frac{1}{2}} - 1 | \approx 0.414 $$
and taking a fourth root would allow:
$$ | 2^{\frac{1}{4}} - 1 | \approx 0.189 $$
So computing $\ln 2^{\frac{1}{4}}$ by the power series expansion will converge more rapidly than for $\ln(2/e)$ and avoids computing $2/e$ in advance by taking a couple of square roots (to six digits of precision).
Of course they have calculators (in software) that run circles around us chemical computers these days!

Now if we have to compute $2/e$ from scratch, it is not a terribly difficult task.  Certainly it does not require "more than two mil[l]ion" terms in the exponential series.
In fact we can compute $e^{-1}$ using the Taylor series expansion:
$$ e^{-1} = \sum_{k=0}^\infty \frac{(-1)^k}{k!} $$
and by the Alternating Series Theorem, the error of truncation after the $k$th term of this series is less than the (absolute value of the) $k+1$st term.  Summing through $k=10$ should get you almost the desired precision; using $k=11$ should safeguard your accuracy.
It's a bit of work, but with care it will easily fit on one sheet of paper.  Then:
$$ 2/e = e^{-1} + e^{-1} $$
and we are ready to use the power series expansion of logarithm (on $(2/e) - 1$).
The other detail to be fastidious about is how many terms of the logarithm's power series you will actually need.  Consult Taylor's Theorem with Remainder for details of how many terms are needed to shrink the error of truncation, and bear in mind that if you have approximated $2/e$, this error must be combined with the error of truncation in approximating $\ln(2/e)$ to get a true estimate of accuracy.
