summation notion beginning at n=1...
[(-1)^n]/n
No calculator use please. Any way to figure where it converges?
Yes, it is $\displaystyle -\ln(1+x)$. Here's why:
Start with the alternating geometric series: $\displaystyle \frac{1}{1+x}=1-x+x^2-x^3+...$
Now make each side negative: $\displaystyle -\frac{1}{1+x}=-1+x-x^2+x^3-...$
Integrate both sides: $\displaystyle -\ln(1+x)=-x+\frac{x^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}$
So $\displaystyle -\ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^n}{n}x^n$
Plugging in $\displaystyle x=1$ gives you $\displaystyle \boxed{\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=-\ln(2)}$