The harmonic alternative Two

we will group the series as follows $A = \frac{1}{1} - \frac{1}{4} + \frac{1}{7} .....$ $B = \frac{1}{2} - \frac{1}{5} + \frac{1}{8} ....$ $C = \frac{1}{3} - \frac{1}{6} + \frac{1}{9} ....$

for the series C we can take 1/3 as common to get $C = \frac{1}{3} ( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} ........... )$ $C = \frac{1}{3} (ln2 )$ for the other two series we use the general result $\int_0^1 \frac{x^a}{1+x^d} dx = \frac{1}{a+1} - \frac{1}{a+1+d} + \frac{1}{a+1+2d} - .........$ $A+B = \int_0^1 \frac{1}{1+x^3} dx + \int_0^1 \frac{x}{1+x^3} dx$ which gives $= \int_0^1 \frac{1+x}{1+x^3} dx$ $= \int_0^1 \frac{1}{x^2-x+1}$ $= \int_0^1 \frac{1}{(x-\frac{1}{2})^2 + (\frac{\sqrt3}{2})^2 } dx$ $= \frac{2}{\sqrt3} tan^{-1}(\frac{2x-1}{\sqrt3}) |^{1}_0$ $= \frac{ 2 \pi \sqrt3}{9}$

hence the total result is $S = A + B + C$ $S = \frac{ 2 \pi \sqrt3}{9} + \frac{1}{3} ln2$

which is 1.44 correct to 3 decimal places.