# Proof of Convergence

## 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.

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Ray Vickson
Homework Helper
Dearly Missed

## 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 $3n < 3 n^2$).

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.

Homework Helper
A little simpler:
$$\dfrac n{n^2+3n} = \dfrac{n}{n(n+3)} = \dfrac{1}{n+3} \ge \dfrac 1 {2n}$$
for $n \ge 3$