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Proof of Convergence

  • Thread starter Calu
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  • #1
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Homework Statement



I have been asked to prove the convergence or otherwise of ∑n=1 n/(3n + n2).

In the example solution, with the aim to prove divergence by comparison with the Harmonic Series, the lecturer has stated that n/(3n + n2) ≥ n/(4n2) = 1/4n and which diverges to +∞.

I was wondering how to arrive at the decision to write n/(3n + n2) ≥ n/(4n2) came from, and how I would arrive at a similar inequality in further examples.
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



I have been asked to prove the convergence or otherwise of ∑n=1 n/(3n + n2).

In the example solution, with the aim to prove divergence by comparison with the Harmonic Series, the lecturer has stated that n/(3n + n2) ≥ n/(4n2) = 1/4n and which diverges to +∞.

I was wondering how to arrive at the decision to write n/(3n + n2) ≥ n/(4n2) came from, and how I would arrive at a similar inequality in further examples.
For ##a,b,c,> 0## you have ##a/b > a/c## if ##b < c##. In other words, if you make the denominator bigger you make the fraction smaller. Of course, in this case you have ##3n + n^2 < 4 n^2## for ##n > 1## (when ##n < n^2##, so ##3n < 3 n^2##).
 
  • #3
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For ##a,b,c,> 0## you have ##a/b > a/c## if ##b < c##. In other words, if you make the denominator bigger you make the fraction smaller. Of course, in this case you have ##3n + n^2 < 4 n^2## for ##n > 1## (when ##n < n^2##, so ##3n < 3 n^2##).
I see, thank you very much.
 
  • #4
statdad
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A little simpler:
[tex]
\dfrac n{n^2+3n} = \dfrac{n}{n(n+3)} = \dfrac{1}{n+3} \ge \dfrac 1 {2n}
[/tex]
for [itex] n \ge 3 [/itex]
 

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