Quick question

1. Mar 2, 2005

hola

Hey. This has been bugging me for a long time:
why does summation from n=1 to infinity of (-1)^n or i^n or 1/n or -1/n not converge, because summation from n=1 to infinity of 1/n^2 conveges. Don't the terms in (-1)^n or i^n or 1/n or -1/n(-1)^n tend to 0?

2. Mar 2, 2005

t!m

While tending towards 0 is a requirement for convergence, it does not guarantee convergence. The most classic example, which you cited, would be the harmonic series, 1/n which does have a limit of 0 but still diverges. Though a limit of anything other than 0 would guarantee divergence.

3. Mar 3, 2005

Galileo

$(-1)^n$ and $\quad i^n$ do not tend to zero, so the associate series is not convergent.

4. Mar 3, 2005

dextercioby

Well,the trick is with the series
$$\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k}$$.
IIRC it converges to
$$\ln 2$$.

Daniel.

5. Mar 3, 2005

saltydog

Hum, interesting. Think I might try to show that. From your post the other day, seems Euler Gamma functions may be involved.

6. Mar 3, 2005

Muzza

I think it's just a matter of using the Taylor series for the logarithm.

7. Mar 3, 2005

dextercioby

You can start from this series
$$\frac{1}{1+x}=1-x+x^{2}-x^{3}+x^{4}-...$$

and integrate it term by term and you'll have to put x=1 in the new series to find the formula which i've hinted.

Daniel.

8. Mar 3, 2005

saltydog

Thanks, I understand now (should know that I know). Actually I'd like to understand this one:

$$\sum_{k=1}^{n}\frac{1}{2k}=\frac{1}{2}\psi(n+1)+\frac{1}{2}\gamma$$

I'll do some work on it and if I can't figure it out, I'll make a separate post in the general math and well, ask for help.