How to begin this problem to find convergence/divergence?

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jaemnhyuck

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I have to find whether the above series converges absolutely, converges conditionally, or diverges. However, I'm not sure where to begin. We were told to always start with the divergence test but I don't really know how to take the limit of this. My guess is that the limit is infinity because of the n^(n^2) term getting much bigger than the rest of the terms. If so, how do I show that? Thanks!
 
jaemnhyuck said:
View attachment 20364

I have to find whether the above series converges absolutely, converges conditionally, or diverges. However, I'm not sure where to begin. We were told to always start with the divergence test but I don't really know how to take the limit of this. My guess is that the limit is infinity because of the n^(n^2) term getting much bigger than the rest of the terms. If so, how do I show that? Thanks!
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Your intuition is correct. Try underestimating the terms by making the numerator smaller and the denominator larger while making it simpler to work with, but still not going to zero. To get you started you could note that [MATH]n^{n^2} + n^{\frac 1 n} + \sin\frac 1 n \ge n^{n^2}[/MATH]. Now, for the denominator...
 
Could you explain why we're trying to make the denominator larger and the numerator smaller? Wouldn't that mean the limit goes to 0?
 
jaemnhyuck said:
Could you explain why we're trying to make the denominator larger and the numerator smaller? Wouldn't that mean the limit goes to 0?
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Maybe you are correct about the limit going to 0. It sounds like you are making a guess at that. Can you try to calculate that limit and see what you get. Post back showing your work. If it happens that this new limit diverges that the larger function (the original one) will also diverge by the direct comparison test.
 
By the way, you said that nn^2 dominates the denominator so why are you now saying that the new limit will go to zero?
 
jaemnhyuck said:
Could you explain why we're trying to make the denominator larger and the numerator smaller? Wouldn't that mean the limit goes to 0?
Click to expand...
If I have a quantity Q that I want to show goes to infinity, and I find something smaller than Q that goes to infinity, wouldn't that do it? Your series will diverge if the nth term doesn't go to zero. If you can show the nth term goes to infinity, as you suggest in your OP, it certainly doesn't go to zero.
 
You are summing an infinite number of terms so the terms must go to zero or else the sum cannot be finite. You can see that for n large,...
 
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