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

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.

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

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?

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
Homework Helper
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
Homework Helper
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.

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