Absolute and conditional convergence of series (1 + x/n)^n^2

[looserlama]

Homework Statement

\sum from n=1 to n=\infty (1 + \frac{x}{n})ⁿ²

Determine the values of x for which the series converges absolutely, converges conditionally and diverges.

The Attempt at a Solution

So i tried using the root test for the absolute value of (1 + \frac{x}{n})ⁿ², but it was inconclusive.

So then I tried using the ratio test, but I need to simplify this:

|1 + \frac{x}{n+1}|⁽ⁿ⁺¹⁾² / |1 + \frac{x}{n}|ⁿ²

and I have no idea how to simplify this.

I don't know any other tests that we could use for this, so I'm just pretty stumped.

Any help would be great.

 

[Dick]

I don't think the root test is all that inconclusive. What is the nth root of your expression?

 

[looserlama]

So what I got was:

nth-root(|1 + \frac{x}{n}|^{n2 ) = [nth-root(|1 + \frac{x}{n}|n )]2 =} |1 + \frac{x}{n}|² = (1+ \frac{x}{n})²

So the limit of this as n goes to infinity would be 1, and that means the test is inconclusive no?

Unless I did something wrong?

 

[looserlama]

Oh wait, that's not right.

nth-root(|1 + x/n|ⁿ²) = |1 + x/n|^n2/n = |1 + x/n|ⁿ

But what does that go to when n goes to infinity?

I don't think I know how to compute that?

 

[Dick]

So what I got was:

nth-root(|1 + \frac{x}{n}|^{n2 ) = [nth-root(|1 + \frac{x}{n}|n )]2 =} |1 + \frac{x}{n}|² = (1+ \frac{x}{n})²

So the limit of this as n goes to infinity would be 1, and that means the test is inconclusive no?

Unless I did something wrong?

Yes, it's completely wrong. Your rules of exponents are pretty messed up. The nth root of a is a^(1/n). What's (a^(n^2))^(1/n)?

 

[Dick]

Oh wait, that's not right.

nth-root(|1 + x/n|ⁿ²) = |1 + x/n|^n2/n = |1 + x/n|ⁿ

But what does that go to when n goes to infinity?

I don't think I know how to compute that?

I think you've probably seen that limit before. It's related to the exponential function. If not then you can work it by taking the log and using l'Hopital's theorem.

 

[looserlama]

I think you've probably seen that limit before. It's related to the exponential function. If not then you can work it by taking the log and using l'Hopital's theorem.

Oh wow, I feel like an idiot now, haha.

Yea I know that limit n\rightarrow\infty(1 + x/n)ⁿ = e^x, but (1 + x/n)ⁿ doesn't necessarily equal |1 + x/n|ⁿ does it? What if x is a large negative number?

I showed it using logarithms as you said, but I was just wondering about that?

 

[Dick]

Oh wow, I feel like an idiot now, haha.

Yea I know that limit n\rightarrow\infty(1 + x/n)ⁿ = e^x, but (1 + x/n)ⁿ doesn't necessarily equal |1 + x/n|ⁿ does it? What if x is a large negative number?

I showed it using logarithms as you said, but I was just wondering about that?

If x is a large negative number, then if n>|x|, 1+x/n is still a positive number. When you are doing limits you only have to worry about n very large.