Sum of series with conditional convergence

Question

Sorry for this question, but for some reason I'm stuck on this for few hours already. Before I solved more complex ( I think ) problems, but can't solve this. The only thing I know that this series converge conditionally, but that is just thing you can tell immediately.

${ 1 + \frac{1}{2} + \frac{1}{3} - \frac{1}{4} - \frac{1}{5} - \frac{1}{6} + \frac{1}{7} +\frac{1}{8} + \frac{1}{9} - \frac{1}{10} - \frac{1}{11} - \frac{1}{12} + ... }$

I will be really grateful if someone at least give me hint how to do it.

P.S. In task book where I got this problem it was said I need to use harmonic series partial sum formula, but I cannot find the way to use it here.

Answer 1

HINT:

The series of partial sums $S_N$ is

$$ S_N=\sum_{n=1}^N \left(\frac1{6n-5}+\frac1{6n-4}+\frac1{6n-3}-\frac1{6n-2}-\frac1{6n-1}-\frac1{6n}\right) $$

Answer 2

There's a standard way to do these things. Note that $${x \over 1 + x^3} = x(\sum_{n=0}^{\infty} (-x^3)^n)$$ $$=\sum_{n=0}^{\infty} (-1)^n x^{3n + 1}$$ As a result (I'll leave the rigorousness for you to ponder on), $$\int_0^1 {x \over 1 + x^3} = \sum_{n=0}^{\infty} \int_0^1 (-1)^n x^{3n + 1}$$ $$= \sum_{n=0}^{\infty} {(-1)^{n} \over 3n + 2}$$ $$= {1 \over 2} - { 1 \over 5}+ { 1\over 8} -...$$ This gives you one third of the series, and you can do similar integrals for the other two portions of it. If you evaluate these integrals and add the results you'll get the answer.

Your work can be reduced if you recognize one of the other two thirds of the series as a multiple of the alternating series for $\ln 2$. The sum of the integrals for the remaining two thirds will then simplify a bit.