Does This Sequence Converge or Diverge?

ironman
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Homework Statement


Show if this sequence (with n=1 to infinity) diverge or converge

Homework Equations


[/B]
CodeCogsEqn-5.gif


The Attempt at a Solution



If I use the Limit Comparison Test:

compare with
CodeCogsEqn-6.gif
so you get
CodeCogsEqn-7.gif
that equals
CodeCogsEqn-8.gif
lim n -> inf
CodeCogsEqn-8.gif
=> inf.

Can I use the Test like this? What does this tell me? (If anything at all...)
 
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I suggest doing limit comparison with ##(2/3)^n## instead.
 
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Orodruin said:
I suggest doing limit comparison with ##(2/3)^n## instead.

Ah the limit is now 1 .
So:
Limit < infinity.
And ##(2/3)^n## converges.
So the original function must converge as well.
 
ironman said:
Ah the limit is now 1 .
So:
Limit < infinity.
And ##(2/3)^n## converges.
So the original function must converge as well.

Right, and a direct comparison is easier. First, for ##a > 1## the function ##a^n## always swamps ##n^k## for any fixed power ##k##, meaning that for large enough ##n## we always have ##a^n > n^k##. So, if
t_n = \frac{2^n + n^2}{3^n + n^3}
we have that the numerator is ##2^n + n^2 < 2^n + 2^n = 2 \cdot 2^n## for all ##n > n_0## (where you can even find ##n_0## if you want). The denominator is ##3^n + n^3 > 3^n##, so ##0 < t_n < \frac{2 \cdot 2^n}{3^n} = 2(2/3)^n##, and now you are almost done.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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