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

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MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,684
St. Augustine, FL.
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.
 
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MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,684
St. Augustine, FL.
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.