The Royal Primate's question at Yahoo Answers regarding convergence of a series

MarkFL

Pessimist Singularitarian
Staff member
Here is the question:

The Royal Primate at Yahoo Answers said:
Need help determining whether a series converges or diverges?

I have the following series:

Sigma (from n=1 to infinity): (n^3 + n^2)/(5n^4 + 3)

and I need to determine if it converges or diverges. I am given that the series diverges by the Limit Comparison Test with the Harmonic Series but I am not sure how to do the work to get there.

Thanks!
Update : I need to find the prototype of this series, would that just be (n^3)/(5n^4) = (1/5n)?
I have posted a link there to this thread so the OP can view my work.

MarkFL

Pessimist Singularitarian
Staff member
Hello The Royal Primate,

We are given to determine whether the followin series converges:

$$\displaystyle S=\sum_{n=1}^{\infty}\frac{n^3+n^2}{5n^4+3}$$

If we observe that for all natural numbers $n$, we have:

$$\displaystyle 5n^3>3$$

Now add $5n^4$ to the both sides:

$$\displaystyle 5n^4+5n^3>5n^4+3$$

Factor the left side:

$$\displaystyle 5n\left(n^3+n^2\right)>5n^4+3$$

Divide through by $$\displaystyle 5n\left(5n^4+3\right)>0$$:

$$\displaystyle \frac{n^3+n^2}{5n^4+3}>\frac{1}{5n}$$

Now, since we know the series:

$$\displaystyle \sum_{n=1}^{\infty}\frac{1}{5n}=\frac{1}{5}\sum_{n=1}^{\infty}\frac{1}{n}$$

is a divergent harmonic series, we therefore know that:

$$\displaystyle S=\sum_{n=1}^{\infty}\frac{n^3+n^2}{5n^4+3}$$

diverges as well since every term in the generated sequence is greater than those in the known divergent series. And so by the direct comparison test we know that $S$ diverges.