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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


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

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I don't see how your approach was supposed to work. Regarding how to actually calculate the sum: – M.B. Jun 15 '12 at 12:36
You cannot re-arrange, or split-up the series in two parts, or use any of that: your series is not absolutely convergent and so any "rearranging" can seriously screw up the sum. – Willie Wong Jun 15 '12 at 12:40
up vote 6 down vote accepted

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 $\log 2$.

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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.

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$\left[ - \sum\limits_{k = 0}^\infty {\ln \left( {(1 - q)(x{q^k} - \alpha ) + 3q - 1} \right)} + \sum\limits_{k = 0}^\infty {\ln \left( {\frac{{(1 - q)}}{2}(x{q^k} - (\alpha + 1))} \right)} \right]$

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Welcome to math.SE: could you please expand a bit your answer? – A.P. Apr 23 '13 at 16:56

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