# Is this series divergent or convergent?

• TheoEndre
In summary: The conversation is about determining whether the series ##\sum_{n=1}^{\infty }1+(-1)^{n+1} i^{2n}## is divergent or convergent. The person asking the question attempted to use the divergence test by taking the limit as ##n## approaches ##{\infty }##, but was confused by the alternating nature of the terms. Another person suggested looking at two cases: when ##n## is even and when it is odd. It was concluded that each term is equal to 0, so the series is divergent. However, someone pointed out that the terms can be simplified and the series is actually convergent. Overall, the conversation was focused on simpl
TheoEndre

## Homework Statement

##\sum_{n=1}^{\infty }1+(-1)^{n+1} i^{2n}##
Is this series divergent or convergent?

## Homework Equations

3. The Attempt at a Solution [/B]
I tried using the divergent test by taking the limit as ##n## approaches ##{\infty }##, but both ##i^{2n}## and ##(-1)^{n+1}## will alternate from ##-1## to ##1##, while when I write the first terms, I get ##0## and that's what made me confused.

TheoEndre said:

## Homework Statement

##\sum_{n=1}^{\infty }1+(-1)^{n+1} i^{2n}##
Is this series divergent or convergent?

## Homework Equations

3. The Attempt at a Solution [/B]
I tried using the divergent test by taking the limit as ##n## approaches ##{\infty }##, but both ##i^{2n}## and ##(-1)^{n+1}## will alternate from ##-1## to ##1##, while when I write the first terms, I get ##0## and that's what made me confused.
Using the n-th term test for divergence, the problem boils down to finding ##\lim_{n \to \infty}\left(1 + (-1)^{n + 1} \cdot i^{2n}\right) = 1 + \lim_{n \to \infty} \left((-1)^{n + 1} \cdot i^{2n}\right)##
It's true that the two factors in the last limit oscillate, but can you determine whether the last limit actually exists? Look at two cases: 1) n is even, and 2) n is odd.

I suggest working a bit with the expression inside the sum ...

fresh_42
Mark44 said:
Using the n-th term test for divergence, the problem boils down to finding ##\lim_{n \to \infty}\left(1 + (-1)^{n + 1} \cdot i^{2n}\right) = 1 + \lim_{n \to \infty} \left((-1)^{n + 1} \cdot i^{2n}\right)##
It's true that the two factors in the last limit oscillate, but can you determine whether the last limit actually exists? Look at two cases: 1) n is even, and 2) n is odd.
I think I get it. For both cases the result will be ##-1##, so last limit would be ##-1## and we'll end up with ##0##, correct?

TheoEndre said:
I think I get it. For both cases the result will be ##-1##, so last limit would be ##-1## and we'll end up with ##0##, correct?
You'll end up with 0 for the last limit I showed, yes. So ##\lim_{n \to \infty}\left(1 + (-1)^{n + 1} \cdot i^{2n}\right) = 1 + 0 = 1##. Can you conclude anything from the n-th term test for divergence, or will you need to use some other test?

Mark44 said:
You'll end up with 0 for the last limit I showed, yes. So ##\lim_{n \to \infty}\left(1 + (-1)^{n + 1} \cdot i^{2n}\right) = 1 + 0 = 1##. Can you conclude anything from the n-th term test for divergence, or will you need to use some other test?
This is not correct. The second limit is the limit of a number of modulus one. It cannot possibly be zero.

TheoEndre said:
I think I get it. For both cases the result will be ##-1##, so last limit would be ##-1## and we'll end up with ##0##, correct?
You actually do not need any limits at all. Each term is identically zero.

Mark44 said:
You'll end up with 0 for the last limit I showed, yes. So ##\lim_{n \to \infty}\left(1 + (-1)^{n + 1} \cdot i^{2n}\right) = 1 + 0 = 1##. Can you conclude anything from the n-th term test for divergence, or will you need to use some other test?
Actually, I meant this ##\lim_{n \to \infty} \left((-1)^{n + 1} \cdot i^{2n}\right)## to equal ##-1## which means ##\lim_{n \to \infty}\left(1 + (-1)^{n + 1} \cdot i^{2n}\right) ## will equal zero.

Orodruin said:
You actually do not need any limits at all. Each term is identically zero.
I thought about that too. But the divergence test of it made me a bit confused.

TheoEndre said:
I thought about that too. But the divergence test of it made me a bit confused.
Before you consider the sum, consider the summands. ##a_n=1+(-1)^{n+1}i^{2n}## can be simplified a lot. After done that for an arbitrary ##a_n##, only then start to think about ##\sum_n a_n##.

Mark44 said:
You'll end up with 0 for the last limit I showed, yes. So ##\lim_{n \to \infty}\left(1 + (-1)^{n + 1} \cdot i^{2n}\right) = 1 + 0 = 1##. Can you conclude anything from the n-th term test for divergence, or will you need to use some other test?
Orodruin said:
This is not correct. The second limit is the limit of a number of modulus one. It cannot possibly be zero.
I am not seeing it.
Regarding the product, if n is an even integer (=2m), ##(-1)^{2m + 1} = -1## and ##i^{2(2m} = 1## with a product of -1.If n is an odd integer (= 2m + 1), ##(-1)^{2m + 2} = 1## and ##i^{2(2m + 1)} = i^{4m + 2} = -1##, so the product is also -1. Where is the flaw in this logic?

Mark44 said:
I am not seeing it.
Regarding the product, if n is an even integer (=2m), ##(-1)^{2m + 1} = -1## and ##i^{2(2m} = 1## with a product of -1.If n is an odd integer (= 2m + 1), ##(-1)^{2m + 2} = 1## and ##i^{2(2m + 1)} = i^{4m + 2} = -1##, so the product is also -1. Where is the flaw in this logic?
No flaw. The product is constantly ##-1##, so ##1+(-1)=0##.

Mark44 said:
I am not seeing it.
Regarding the product, if n is an even integer (=2m), ##(-1)^{2m + 1} = -1## and ##i^{2(2m} = 1## with a product of -1.If n is an odd integer (= 2m + 1), ##(-1)^{2m + 2} = 1## and ##i^{2(2m + 1)} = i^{4m + 2} = -1##, so the product is also -1. Where is the flaw in this logic?
No flaw in that, but you seemed to be saying it was zero, not -1.

Orodruin said:
No flaw in that, but you seemed to be saying it was zero, not -1.
Doh! I got the hard part right, but fouled up on 1 + -1.

fresh_42
fresh_42 said:
Before you consider the sum, consider the summands. ##a_n=1+(-1)^{n+1}i^{2n}## can be simplified a lot. After done that for an arbitrary ##a_n##, only then start to think about ##\sum_n a_n##.
I tried simplifying and ended up with zero So the series converges and the sum equal to ##0##, correct?

TheoEndre said:
I tried simplifying and ended up with zero So the series converges and the sum equal to ##0##, correct?
Yes, you can add as many zeroes as you want, you won't get very far. Formally an induction would do, in case anyone comes up with a nonsense like ##\infty \cdot 0##.

fresh_42 said:
Yes, you can add as many zeroes as you want, you won't get very far. Formally an induction would do, in case anyone comes up with a nonsense like ##\infty \cdot 0##.
I think ##\infty \cdot 0## is interesting
Thank very much for your help @fresh_42, @Mark44 and @Orodruin

## 1. What does it mean for a series to be divergent or convergent?

A divergent series is one in which the sum of all its terms approaches infinity as the number of terms increases. In contrast, a convergent series is one in which the sum of its terms approaches a finite value as the number of terms increases.

## 2. How can I determine if a series is divergent or convergent?

There are several tests that can be used to determine the convergence or divergence of a series, such as the ratio test, the root test, and the integral test. These tests involve examining the behavior of the terms in the series and can be applied to different types of series, such as geometric, harmonic, and power series.

## 3. Can a series be both divergent and convergent?

No, a series can only be either divergent or convergent. If the series approaches infinity, it is divergent, and if it approaches a finite value, it is convergent. However, some series may have both divergent and convergent parts, in which case we say that the series is conditionally convergent.

## 4. What happens if a series is divergent?

If a series is divergent, its sum is said to be infinite, meaning that it does not have a finite value. In other words, the series does not have a sum and cannot be evaluated.

## 5. Why is it important to determine if a series is divergent or convergent?

Knowing whether a series is divergent or convergent is crucial in many areas of mathematics and science. For example, in calculus, the convergence or divergence of a series can determine the convergence or divergence of a corresponding integral. In physics, series are often used to represent physical phenomena, and knowing whether they are convergent or divergent can provide important insights into these phenomena.

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