<s>[INST] Summation of a series. I encountered this problem in Physics before i knew about a thing called Taylor Polynomials My problem was that i had to sum this series :
$$\sum^\infty_{n=1}\frac{(-1)^{n+1}}{n}$$
basically $$1,-\frac{1}{2},\frac{1}{3},-\frac{1}{4},\frac{1}{5},-\frac{1}{6},\frac{1}{7}.....$$
So now i know that there is something called a taylor polynomial that says that 
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\frac{x^7}{7}....$$
So the above summation boils down to $\ln 2$.
What if i never knew the exansion then how would I calculate it?
Earlier I tried solving it like so ,
divide it into two different sets i.e.
$$\text{1 and $\dfrac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13} \ldots$ and $-\dfrac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\ldots$}$$
I said Hey! the first set would contain stuff like, $$\frac{1}{3^n},\frac{1}{5^n},\ldots$$ each of them would probably be reduced to a sum like so $$\sum^\infty_{n=1}\frac1{a^n}=\frac1{a-1}$$ and further become $$\sum^\infty_{a=3}\frac1{a-1}$$ which would subtract all the numbers in the other even set giving 1 as the answer which is wrong . 
Where did I go wrong and how could I proceed even without knowing Taylor polynomials
 [/INST] $\left[ - \sum\limits_{k = 0}^\infty  {\ln \left( {(1 - q)(x{q^k} - \alpha ) + 3q - 1} \right)}  + \sum\limits_{k = 0}^\infty  {\ln \left( {\frac{{(1 - q)}}{2}(x{q^k} - (\alpha  + 1))} \right)} \right]$
</s>