Proof of Convergence: ∑∞n=1 n/(3n + n2)

[Calu]

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.

 

[Ray Vickson]

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##).

 

[Calu]

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.

 

[statdad]

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]