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Root test fails, now what?

  • #1

Homework Statement



determine whether the series (1-1/n^(1/3))^n converge or diverge

Homework Equations


all the testing procedure



The Attempt at a Solution


So I did the root test first, but the limit on the inside is 1. I then tried the ratio test but then when I tried taking the limit, I ended up with e^(-infinity+infinity). Finally, I tried the test for divergence on the original series, but i turned out be zero so it might or might not converge. Now I'm totally stuck.
 

Answers and Replies

  • #2
D H
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Did you try the comparison test?
 
  • #3
33,497
5,188

Homework Statement



determine whether the series (1-1/n^(1/3))^n converge or diverge

Homework Equations


all the testing procedure



The Attempt at a Solution


So I did the root test first, but the limit on the inside is 1. I then tried the ratio test but then when I tried taking the limit, I ended up with e^(-infinity+infinity).
You should never end up with something like this ( -∞ + ∞), as it is indeterminate. I agree with what you got on the root test, so that's not a useful test, and the n-th term test seems to give a value of 0, so it doesn't tell us anything.

I would take another look at the ratio test to see what the limit actually is.


Finally, I tried the test for divergence on the original series, but i turned out be zero so it might or might not converge. Now I'm totally stuck.
 
  • #4
Ray Vickson
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Homework Statement



determine whether the series (1-1/n^(1/3))^n converge or diverge

Homework Equations


all the testing procedure



The Attempt at a Solution


So I did the root test first, but the limit on the inside is 1. I then tried the ratio test but then when I tried taking the limit, I ended up with e^(-infinity+infinity). Finally, I tried the test for divergence on the original series, but i turned out be zero so it might or might not converge. Now I'm totally stuck.
If
[tex]t_n = \left(1 - \frac{1}{n^{1/3}}\right)^n,[/tex]
you can look at ##L_n = \ln(t_n)## and use the series expansion of ##\ln (1-x)## for small ##x = 1/n^{1/3}##, to get the behavior of ##t_n## for large n. In fact, you can even get a simple upper bound ##u_n##, so that ## 0 < t_n < u_n##, and ##\sum u_n## is easy to analyze.
 
  • #5
33,497
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The OP might not have been exposed to series yet...
 
  • #6
Ray Vickson
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The OP might not have been exposed to series yet...
Series are not really needed here; it is enough to know that 1-x < exp(-x) for x > 0, so that ln(1-x) < -x for small x > 0.

Anyway, if the OP has seen series the method could be useful; if he/she has not yet seen series it is perhaps not useful. However, I have no way of knowing what the OP has, or has not, seen.
 
Last edited:
  • #7
D H
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I'm not seeing how that would help. If it's working toward the ratio test, the ratio of consecutive terms approaches 1 as n→∞, so that test is inconclusive.

There is a very simple series that does converge that looks somewhat like this.
 

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