# Test Convergence: Sum of i^n/n | Chris Maness

• kq6up
In summary, the homework statement is asking if a series of numbers called \sum_{n=1}^{\infty}{ i^n/n } converges. The test to see if it does is to see if it can be split into real and imaginary parts and both series converge. However, if you rearrange the terms in the series, the result may not be the same as the original series.
kq6up

## Homework Statement

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

See above.

## 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

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.

jbunniii said:
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.

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

kq6up said:
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.
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.

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

## 1. What is the formula for the sum of i^n/n?

The formula for the sum of i^n/n is: ∑i^n/n = 1/1 + 2^1/1 + 3^1/1 + 4^1/1 + ... + n^1/1

## 2. How is this formula related to test convergence?

This formula is related to test convergence because it is used to determine if a series converges or diverges. The value of the sum of i^n/n is compared to a known series that either converges or diverges. If the value is greater than the known series, then the series diverges. If the value is less than the known series, then the series converges.

## 3. What is the significance of using i^n/n in this formula?

The significance of using i^n/n in this formula is that it allows for a more versatile and complex series to be tested for convergence. The use of i^n/n allows for the values of n to increase at a faster rate, creating a more challenging series to evaluate.

## 4. How is this formula used in real-world applications?

This formula is used in real-world applications to analyze and predict the behavior of various systems and processes. It is commonly used in physics, engineering, and economics to model and understand complex systems.

## 5. Are there any limitations to using this formula?

Yes, there are limitations to using this formula. It can only be used to test for convergence or divergence of a series, not to determine the exact value of the series. Additionally, it may not work for all types of series, such as alternating series. It is important to consider the conditions for convergence and divergence when using this formula.

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