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
 A: I find Norbert's solution more appealing if you run it backwards. 
You're trying to evaluate $$1-{1\over2}+{1\over3}-{1\over4}+\cdots$$ Let $$f(x)=x-{1\over2}x^2+{1\over3}x^3-{1\over4}x^4+\cdots$$ Then we want $f(1)$. So, how can we find a simple formula for $f(x)$? Differentiate it: $$f'(x)=1-x+x^2-x^3+\cdots$$ Recognize this as a geometric series, first term $1$, constant ratio $-x$, so sum is $$f'(x)={1\over1+x}$$ Having differentiated, now antidifferentiate to get $$f(x)=\int{1\over1+x}\,dx=\log(1+x)+C$$ But what is $C$? Well, from the origial formula for $f$, we see $f(0)=0$, so that forces $C=0$, so $f(x)=\log(1+x)$, so $f(1)$, which is what we wanted, is $\log 2$. 
A: I assume that you know basic integration and formula of sum of infinite geometric series. Recall that
$$
\frac{1}{1+t}=\sum\limits_{n=0}^\infty(-t)^n,\quad t\in(-1,1)
$$
Then integrate it over the interval $[0,x]$, whith $-1\leq x\leq 1$ to get 
$$
\ln(1+x)=
\int\limits_{0}^x\frac{1}{1+t}dt=
\int\limits_{0}^x\sum\limits_{n=0}^\infty(-t)^n dt=
\sum\limits_{n=0}^\infty(-1)^n\int\limits_{0}^x t^ndt=
$$
$$
\sum\limits_{n=0}^\infty(-1)^n\frac{x^{n+1}}{n+1}=
\sum\limits_{n=1}^\infty\frac{(-1)^{n+1}x^n}{n}
$$
Now we substitute $x=1$, and obtain
$$
\sum\limits_{n=1}^\infty\frac{(-1)^{n+1}}{n}=\ln 2
$$
Well, this is quite elementary solution, but not rigorous enough - I light-heartedly interchanged summation and integration. But such things won't bother physicists much.
A: $\left[ - \sum\limits_{k = 0}^\infty  {\ln \left( {(1 - q)(x{q^k} - \alpha ) + 3q - 1} \right)}  + \sum\limits_{k = 0}^\infty  {\ln \left( {\frac{{(1 - q)}}{2}(x{q^k} - (\alpha  + 1))} \right)} \right]$
