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

  • Thread starter looserlama
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Homework Statement



[itex]\sum[/itex] from n=1 to n=[itex]\infty[/itex] (1 + [itex]\frac{x}{n}[/itex])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 + [itex]\frac{x}{n}[/itex])n2, but it was inconclusive.

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

|1 + [itex]\frac{x}{n+1}[/itex]|(n+1)2 / |1 + [itex]\frac{x}{n}[/itex]|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.
 

Answers and Replies

  • #2
Dick
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I don't think the root test is all that inconclusive. What is the nth root of your expression?
 
  • #3
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So what I got was:

nth-root(|1 + [itex]\frac{x}{n}[/itex]|n2 ) = [nth-root(|1 + [itex]\frac{x}{n}[/itex]|n )]2 = |1 + [itex]\frac{x}{n}[/itex]|2 = (1+ [itex]\frac{x}{n}[/itex])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?
 
  • #4
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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?
 
  • #5
Dick
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So what I got was:

nth-root(|1 + [itex]\frac{x}{n}[/itex]|n2 ) = [nth-root(|1 + [itex]\frac{x}{n}[/itex]|n )]2 = |1 + [itex]\frac{x}{n}[/itex]|2 = (1+ [itex]\frac{x}{n}[/itex])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)?
 
  • #6
Dick
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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.
 
  • #7
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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[itex]\rightarrow\infty[/itex](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?
 
  • #8
Dick
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Oh wow, I feel like an idiot now, haha.

Yea I know that limit n[itex]\rightarrow\infty[/itex](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?
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.
 

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