# Test for Convergence

1. Mar 23, 2014

### kq6up

1. The problem statement, all variables and given/known data

Test to see if $$\sum_{n=1}^{\infty}{ i^n/n }$$ converges.

2. Relevant equations

See above.

3. The attempt at a solution

If I separate this series into real/imag. parts both series diverges by the integral test. However, according to Wolfram Alpha, the series converges to $$\sum_{n=1}^{\infty}{ i^n/n }= -log(1-i)$$

What test would I use to show this converges, and how did I misuse the integral test?

Thanks,
Chris Maness

2. Mar 23, 2014

### jbunniii

The sum of the first four terms is
$$i - 1/2 - i/3 + 1/4 = -1/4 + 2i/3$$
Try finding a general expression for the sum of the four terms starting at, say, $n = 4k+1$ and see what you can conclude.

3. Mar 24, 2014

### kq6up

I split it into real and imaginary sums and used the integral test. However, I am not sure what you mean by the above statement.

Chris KQ6UP

4. Mar 24, 2014

### jbunniii

If you split it into real and imaginary sums then you get two alternating series, so they both converge. Can you show how you applied the integral test?

Also, be very careful about "splitting into real and imaginary sums". In general, if you rearrange the order of the terms in a series, the result may not be the same as the original series. Rearrangement can only be done safely in general if the convergence is absolute, which is not the case here.

Indeed, a very interesting theorem is the Riemann rearrangement theorem: if you start with a conditionally convergent series, it's possible to rearrange the terms to obtain any result you want: you can make it converge to any limit $L$, or you can make it diverge to $+\infty$ or $-\infty$. So if you rearrange the series as you are doing, you will have to carefully justify why doing so does not change the result.

5. Mar 24, 2014

### kq6up

Ah, I see my mistake now. I assumed the even powers would always be negative, and the odds always positive. I didn't think that they would be alternating. Yes, they would converge by the alternating series test.

Thanks,
Chris