Difficult Infinite Series

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Homework Statement


Determine whether the following series converges absolutely, converges conditionally, or diverges.
[tex]\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3 - ln(n)}[/tex]


Homework Equations


The assortment of different tests.

The Attempt at a Solution


Okay, first of all, I tried using Alternating Series Test. This worked and the series satisfied all 3 conditions (decreasing, alternating, and limit as n approaches infinity = 0). This means the series must converge either conditionally or absolutely since I haven't eliminated the possibility of it converging absolutely yet.

So I take the absolute value of the series and if it converges, it's absolutely convergent.
If it diverges, then it's not absolutely convergent. Therefore, that means it's conditionally convergent because I already proved that it must converge in some manner.
So this is
[tex]\sum_{n=1}^{\infty} |\frac{(-1)^n}{n^3 - ln(n)}|[/tex]
[tex]=\sum_{n=1}^{\infty} |\frac{1}{n^3 - ln(n)}|[/tex]

Now how do I find that this series converges or diverges? I tried every test I'm aware of and each was inconclusive. I tried WolframAlpha and it said that the tests were inconclusive, but it gave a number.
Does this mean it absolutely converges? If so, how would I show my work?

I tried using the Direct Comparison test and compared the series with the absolute values to 1000/n^3. However, I'm not entirely sure if 1000/n^3 is greater than the absolute value series for all n terms.
 
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Answers and Replies

  • #2
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It should be fairly easy to show that n^3 - ln(n) > n^2 for all n >= some N, using induction or some other means.

Once you get that, then 1/(n^3 - ln(n)) < 1/(n^2), and it's known that sum(1/(n^2)) is a convergent p-series.
 
  • #3
354
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Okay, cool. Thanks for the example. Do you think that 1000/n^3 works? I tried checking to see if it it's greater for all n using the calculator, and it sure seemed like it, but I can't truly be sure.

In other words, how can I prove that n^3/1000 < or = n^3 - ln(n) for all n or that it isn't? Solving the inequality, I get
[tex]n^3 \ge \frac{1000}{999}ln(n)[/tex]
And here I'm stuck. I'm gonna be mad if I have to use the Lambert W function or whatever. :\

Could I argue that they never intersect and because n^3 is greater than the other function at one point, it must therefore be greater at all n terms? Then again, can I even fool around with the equation by solving for the inequality if I'm not even sure if it's true in the first place?
 
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  • #4
35,019
6,770
Yeah, I think that will work. You can show that n^3 >= (1000/999)ln(n) by showing that n^3 - (1000/999)ln(n) >= 0 for all n larger than some fixed N. One way to do that is to call define f(x) = x^3 - (1000/999)ln(x), find some x0 for which f(x) > 0, and then show that f'(x) > 0 for all x > x0.
 
  • #5
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Thanks!! I absolutely love you guys.

And sorry for bumping this thread up.
 

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