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I am trying to prove that the ratio test is inconclusive for any series of the form $\sum \frac{p(n)}{q(n)}$ where both $p,q$ are polynomial functions.

So for the ratio test to be inconclusive the following needs to occur,

$lim |\frac {a_{n+1}}{a_n}| = 1 $

so then using $\sum \frac{p(n)}{q(n)}$ $lim \frac{p(n+1)}{q(n+1)}\frac{q(n)}{p(n)}.$

From here on I am not quite sure how to proceed, my thinking was for the limit of this function to be one the the highest powers in the numerator and denominator must be equal, and there coefficients must be equal also, as then the limit would go to 1. But I am not sure if that is correct logic, or if there is another part to the explanation.

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Try looking at the asymptotic behavior of $\frac{p(x)}{q(x)}$. If the degree of $p$ is strictly greater than the degree of $q$, then the summand doesn't converge and the sum diverges. If the degree of $p$ is less than or equal to the degree of $q$, then the ratio of the ratio of the polynomials will tend to 1.

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    but how is this form inconclusive for any series? – 2017-02-10
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    If $p(x)$ has leading term $c_{p}x^{k}$ and $q(x)$ has leading term $c_{q}x^{l}$, then the ratio $a_{n+1}/a_{n}$ would tend to $\frac{\frac{c_{p}(n+1)^{k}}{c_{q}(n+1)^{l}}}{\frac{c_{p}n^{k}}{c_{q}n^{l}}}$ as $n$ gets sufficiently large. See what happens when $n$ tends to infinity from there. – 2017-02-10
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    the result would be $(\frac {n}{n+1})^{k-l}$ then as n goes to infinity this limit is 1? so it is inconclusive? – 2017-02-10
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    Well the result would just be 1, and so it is indeed inclusive. – 2017-02-10