Test the series for convergence

  • Thread starter Sweet_GirL
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  • #1
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Main Question or Discussion Point

hello
I have this one:

[tex]\sum_{n=1}^{\infty} \left( 1 - \sqrt[n]{n} \right) [/tex]


mmmmm am sure it will be tested by using one of the comparison tests
but am not getting it
any help?

this is not my homework, actually I finished my college 2 years ago.
 

Answers and Replies

  • #2
179
2
I would start by looking at the behavior of just [itex]\sqrt[n]{n}[/itex] as [itex]n\to\infty[/itex]. Use the techniques of logarithms to find [itex]\lim \sqrt[n]{n}[/itex]. Once you see what [itex]\sqrt[n]{n}[/itex] approaches, you should immediately be able to see how

[tex]
\sum_{n=1}^{\infty} \left( 1 - \sqrt[n]{n} \right)
[/tex]

must behave. If you don't see it, just consider what [itex](1-\sqrt[n]{n})[/itex] must approach, knowing the limit of [itex]\sqrt[n]{n}[/itex].
 
  • #3
24
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well,
[tex]1 - \sqrt[n]{n} \rightarrow 0[/tex] as [tex]n \rightarrow \infty[/tex]

and this will make no sense.
 
  • #4
179
2
well,
[tex]1 - \sqrt[n]{n} \rightarrow 0[/tex] as [tex]n \rightarrow \infty[/tex]

and this will make no sense.
You are quite right; in my zeal, I made a mistake in computing the limit of the n-th root of n and getting 0!
 
  • #5
258
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I might be wrong but try Cauchy's Condensation Test.
[tex]
1 - n^{\frac{1}{n}} = 2^k (1 - 2^{\frac{k}{2^k}})
[/tex]

Which obviously fails the limit test...
 
  • #6
24
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It must be solved by the standart test.

Anyone ?
 
  • #7
179
2
In order to use the CCT your terms need to be positive and non-increasing. This isn't a big deal since we can just negate the sum, and consider the sum starting from n=3.

As for using "standard" tests, what about the integral test? I've only thought as far as:

[tex]\int_3^\infty (\sqrt[x]{x}-1)\ dx \ge \int_3^\infty (\sqrt[x]{3}-1)\ dx = \int_3^\infty (3^{1/x}-1)\ dx[/tex]
 
  • #8
Gib Z
Homework Helper
3,346
4
Theorem: A bounded monotonic sequence converges.
 
  • #9
179
2
Theorem: A bounded monotonic sequence converges.
Yes, but a sequence is very different from a series. Unless you are referring to the partial sums; but this would require that you can bound the partial sums. Can you elaborate on how this is done?
 
  • #10
Gib Z
Homework Helper
3,346
4
The n-th root of n is greater than the n-th root of 1.
 
  • #11
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Still searching for a solution with standard tests ..
 
  • #12
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Try finding a function f(x) such that [tex]\sqrt[x]{x} - 1[/tex] >> f(x) using l'Hôpital's rule, [tex]\sqrt[x]{x} - 1[/tex] > f(x) on [1, ∞), and [itex]\sum_{n=1}^\infty f(n)[/itex] diverges. Then use the comparison test on your series and f(n).
 
Last edited:
  • #13
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I tried that
but its not easy to find that f
and also f must be positive
 
  • #14
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[itex]n^{1/n}-1[/itex] is asymptotic to [itex]\log(n)/n[/itex], so you need to analyze the convergence of [itex]\sum\log(n)/n[/itex]
 
  • #15
179
2
[itex]n^{1/n}-1[/itex] is asymptotic to [itex]\log(n)/n[/itex], so you need to analyze the convergence of [itex]\sum\log(n)/n[/itex]
Ahh, this indeed will do it. Very clever.
 

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