Abel Summation Formula

Question

Out of curiosity I wanted to calculate $$ \sum_{n=1}^{\infty} \frac{\left(-1\right)^{n-1}}{n} = \ln 2 \approx 0.6931471806 $$ using Abel Summation formula $$ \lim_{x\rightarrow\infty} \left\{ \frac{\sum_{n\leq x} \left(-1\right)^{n-1}}{x} + \int_1^x {\rm d}u \, \frac{1-\left(-1\right)^{\lfloor u \rfloor}}{2u^2} \right\} = \int_1^\infty {\rm d}u \, \frac{1-\left(-1\right)^{\lfloor u \rfloor}}{2u^2} \approx 0.6687714032 \, . $$ So any insights in why the result is different?

Answer 1

This last integral is equal to your series, but there's not much of a better way to compute it:

$$\int_1^\infty \frac{1 - (-1)^{\lfloor u \rfloor}}{2u^2} \,du = \int_{ u \geq 1 : \lfloor u \rfloor \mathrm{odd}}\frac{1}{u^2}\,du =\sum_{k = 0}^\infty \int_{2k+1}^{2k+2} \frac{1}{u^2}\,du = \sum_{k = 0}^\infty \left(\frac{1}{2k+1} - \frac{1}{2k+2} \right).$$

How did you calculate your approximation to this last integral?