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Searching for some help with proving that the ratio test is inconclusive for any series that has the form $\sum \frac{p(n)}{q(n)}$ where both $p$ and $q$ are both polynomial functions.

I know that for the ratio test you must adhere to the following criteria,

1) $lim _{n->\infty} |\frac{a_{n+1}}{a_n}| = L < 1$ then the series is convergent.

2) $lim _{n->\infty} |\frac{a_{n+1}}{a_n}| = L > 1$ or $ L = \infty$

3) $lim _{n->\infty} |\frac{a_{n+1}}{a_n}| = L = 1$ then no conclusion can be made.

So how do I show that $\sum \frac{p(n)}{q(n)}$ where both $p$ and $q$ are both polynomial functions has a $lim=1$

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1 Answers 1

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If the degree of $p(n)$ is greater than the degree of $q(n)$ then the term ratio will itself diverge (and so the series must diverge).

If the degree of $p(n)$ is less than the degree of $q(n)$ then the term ratio will go to $L = 0$ (and so the series must converge.

If degree of $p(n)$ is equal to the degree of $q(n)$ then here is how to show that the term ratio approaches $1$ as $n\to\infty$ (so that the naive ratio test is inconclusive):

Let $p(n) = \sum_{k=0}^d p_k n^k$ and $q(n) = \sum_{k=0}^d q_k n^k$; note that the degree $d$ is the same in both cases.

Let $s = \max_k |p_k|$ and let $t = \max_k |q_k|$. Then using $$ \sum_{k=0}^{d-1} n^k = \frac{n^d-1}{n-1} $$ we see that $$ |p_d| n^d - s\frac{n^d-1}{n-1} \leq |p(n)| \leq |p_d| n^d + s\frac{n^d-1}{n-1} \\ |q_d| n^d - t\frac{n^d-1}{n-1} \leq |p(n)| \leq |q_d| n^d + t\frac{n^d-1}{n-1} $$ and the term ratio $r_n = \frac{p(n+1)q(n)}{p(n)q(n+1)}$ obeys $$ \frac{\left( |p_d| (n+1)^d - s\frac{(n+1)^d-1}{n}\right)\left(|q_d| n^d - t\frac{n^d-1}{n-1} \right)}{\left( |q_d| (n+1)^d + t\frac{(n+1)^d-1}{n}\right)\left(|p_d| n^d + s\frac{n^d-1}{n-1} \right)} \leq r_n \leq \frac{\left( |p_d| (n+1)^d + s\frac{(n+1)^d-1}{n}\right)\left(|q_d| n^d + t\frac{n^d-1}{n-1} \right)}{\left( |q_d| (n+1)^d - t\frac{(n+1)^d-1}{n}\right)\left(|p_d| n^d - s\frac{n^d-1}{n-1} \right)} $$ The left hand expression is always $\leq 1$ and the right hand expression is always $\geq 1$. I will show that the limit, as $n\to\infty$, of the left hand expression is $1$; the proof that the right hand expression goes to $1$ is similar, and once you know that these bounding expressions go to $1$, you have $\lim r_n = 1$.

Divide both numerator and denominator by $n^d(n+1)^d$ yielding $$ X_n=\frac{\left( |p_d| - s\frac{(n+1)^d-1}{n(n+1)^d}\right)\left(|q_d| - t\frac{n^d-1}{(n-1)n^d} \right)}{\left( |q_d| + t\frac{(n+1)^d-1}{n(n+1)^d}\right)\left(|p_d| + s\frac{n^d-1}{(n-1)n^d} \right)} = \frac{\left( |p_d| - \frac{s}{n}\left[1-\frac{1}{(n+1)^d}\right]\right)\left(|q_d| - \frac{t}{n-1}\left[1-\frac{1}{n^d} \right]\right)}{\left( |q_d| + \frac{t}{n}\left[1-\frac{1}{(n+1)^d}\right]\right)\left(|p_d| + \frac{s}{n-1}\left[1-\frac{1}{n^d} \right]\right)} $$ $$ X_n = 1- \left( \frac{s}{n|p_d|}+ \frac{t}{(n-1)|q_d|} + \frac{s}{(n-1)|p_d|} +\frac{t}{n|q_d|} \right) + O(n^{-2}) $$So by choosing $n$ large enough we can make that expression arbitrarily close to $1$.

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    thanks for your answer, I will try and learn this now – 2017-02-07