# Thread: How do you find the value of this series?

1. ## How do you find the value of this series?

summation notion beginning at n=1...

[(-1)^n]/n

No calculator use please. Any way to figure where it converges?

2. The power series for $\ln(1+x)$ is $x-\tfrac12x^2 + \tfrac13x^3-\ldots+(-1)^{n-1}\tfrac1nx^n+\ldots$. It converges when |x|<1 but also when x=1 (though this is harder to prove).

3. Uh... what? I don't understand. My series isn't even a power series. I'm not asking whether it converges, but where it converges.

4. Originally Posted by Kaitosan
Uh... what? I don't understand. My series isn't even a power series. I'm not asking whether it converges, but where it converges.
He's actually telling you how to find the sum (which is usually difficult to do in closed form) Try substituting $x = 1$ .

5. Gotcha. Thanks guys.

6. Originally Posted by Opalg
The power series for $\ln(1+x)$ is $x-\tfrac12x^2 + \tfrac13x^3-\ldots+(-1)^{n-1}\tfrac1nx^n+\ldots$. It converges when |x|<1 but also when x=1 (though this is harder to prove).
As I thought about that, I actually think you meant...

$-ln(1+x)$

Since my series start at n=1. Also, I think I can use the expression $ln[1/(1+3x)]$ since it barely converges to $1/3$ lol.

Right?

7. Yes, it is $-\ln(1+x)$. Here's why:

Start with the alternating geometric series: $\frac{1}{1+x}=1-x+x^2-x^3+...$

Now make each side negative: $-\frac{1}{1+x}=-1+x-x^2+x^3-...$

Integrate both sides: $-\ln(1+x)=-x+\frac{x^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}$

So $-\ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^n}{n}x^n$

Plugging in $x=1$ gives you $\boxed{\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=-\ln(2)}$

8. Heh yeah. Thank you redsoxfan! This information is kinda new to me.