Is this series divergent or convergent?

[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.

 

[Mark44]

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.

 

[Orodruin]

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

 

[TheoEndre]

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?

 

[Mark44]

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?

 

[Orodruin]

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.

 

[Orodruin]

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.

 

[TheoEndre]

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.

 

[TheoEndre]

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.

 

[fresh_42]

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]

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.

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?

 

[fresh_42]

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 $1+(-1)=0$.

 

[Orodruin]

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.

 

[Mark44]

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.

 

[TheoEndre]

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?

 

[fresh_42]

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 $\infty \cdot 0$.

 

[TheoEndre]

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