Infinite Series polynomials

Question

This is a homework problem.

Let P and Q be polynomials of degree $k$ and $m$. Assume $Q(n) \neq 0 $ for all $n \in \mathbb{N} $

Prove the series

$\sum_{n=1}^{\infty} {P(n)}/{Q(n)}$

Is convergent if $m \geq k+2$ and divergent if $m \leq k+1 $

From my understanding it would converge when $ m > k$ And therefore i can't figure out how it should be divergent when $m \leq k+1$ because then $m=k+1$ and $m>k$

Answer 1

Write the polynomials as \begin{align} P(x) &= a_kx^k+\dots+a_1x+a_0 \\[4px] Q(x) &= b_mx^n+\dots+b_1x+b_0 \end{align} with $a_k\ne0$ and $b_m\ne0$.

Then you can rewrite $$ \frac{P(n)}{Q(n)}= \frac{n^k}{n^m} \frac{a_k+\dfrac{a_{k-1}}{n}+\dots+\dfrac{a_1}{n^{k-1}}+\dfrac{a_0}{n^k}} {b_m+\dfrac{b_{m-1}}{n}+\dots+\dfrac{b_1}{n^{m-1}}+\dfrac{b_0}{n^m}} $$ Note that the limit of the big fraction is $a_k/b_m=c$. If $c<0$, we can substitute $P(n)$ with $-P(n)$, without affecting the convergence or divergence of the series, because from some point on, all terms of the series have the same sign as $c$.

Thus we can assume $c>0$. Then there exists $N$ such that, for every $n>N$, all terms $P(n)/Q(n)$ have the same sign and $$ 0<\frac{c}{2}\frac{n^k}{n^m}<\frac{P(n)}{Q(n)}<\frac{3c}{2}\frac{n^k}{n^m} $$ Hence the given series converges if and only if $$ \sum_{n}\frac{n^k}{n^m} $$ converges.

If $m\le k$ this series doesn't converge, because the general term doesn't have zero limit. If $m=k+1$ this is the harmonic series, which does not converge.

If $m>k+1$ (that is, $m\ge k+2$) then the series converges by comparison with $1/n^2$.

Answer 2

The idea is that if $m\geq {k+2}$, then $$ \frac{P(n)}{Q(n)}\sim \frac{C}{n^{m-k}} $$ where $f(n)\sim g(n)$ means $f(n)/g(n)\to 1$ as $n\to\infty$ for some constant $C$. The limit comparison test gives us the result. If $m\leq k$ you should be able to show the terms of the series don't go to zero. If $m=k+1$, you should be able to apply limit comparison test by deducing that $$ \frac{P(n)}{Q(n)}\sim \frac{D}{n}. $$ for some constant $D$.

Answer 3

For $n$ big enough, $P(n)/Q(n)$ has always the same sign, so, WLOG, we can assume that we have a series with positive terms(*).

Let $P(n)=p_0+p_1n+\cdots+p_k n^k$. Because $P$ is of degree $k$, $p_k\ne0$.

Let $Q(n)=q_0+q_1n+\cdots+q_m n^m$. Because $Q$ is of degree $m$, $q_m\ne0$.

Now, $$\lim_{n\to\infty}\frac{P(n)/Q(n)}{1/n^{m-k}}= \lim_{n\to\infty}\frac{n^{m-k} P(n)}{Q(n)}=\lim_{n\to\infty}\frac{p_0n^{m-k}+p_1n^{m-k+1}+\cdots+p_kn^m}{q_0+q_1n+\cdots q_mn^m}=\frac{p_0}{q_0}.$$ So your series has the same character as $\sum \frac{1}{n^{m-k}}$ wich is convergente $\iff$ $m-k>1$.

But because $m$ and $k$ are both integers, $m-k>1$ $\iff$ $m-k\ge 2$.

EDIT (*): If $p_k\cdot q_m>0$, then for $n$ big enough, $P(n)/Q(n)>0$ because both $P(n)$ and $Q(n)$ tends simultaneously to $+\infty$ or $-\infty$, when $n\to\infty$.

If $p_k\cdot q_m<0$, then $P(n)\to+\infty$ and $Q(n)\to-\infty$ ($n\to\infty$) or viceversa. Then $P(n)/Q(n)<0$ for $n$ big enough and then you only have to work with $-P(n)/Q(n)$.