Krdytkyu

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 $m>k$

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 $a_kne0$ and $b_mne0$.

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 $mle k$ 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, $mge k+2$) then the series converges by comparison with $1/n^2$.

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 $k$, $p_kne0$.

Let $Q(n)=q_0+q_1n+cdots+q_m n^m$. Because $Q$ is of degree $m$, $q_mne0$.

Now, $$lim_{ntoinfty}frac{P(n)/Q(n)}{1/n^{m-k}}= lim_{ntoinfty}frac{n^{m-k} P(n)}{Q(n)}=lim_{ntoinfty}frac{p_0n^{m-k}+p_1n^{m-k+1}+cdots+p_kn^m}{q_0+q_1n+cdots q_mn^m}=frac{p_0}{q_0}.$$ 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, $m-k>1$ $iff$ $m-kge 2$.

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

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

The idea is that if $mgeq {k+2}$, 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 $ntoinfty$ for some constant $C$. The limit comparison test gives us the result. If $mleq k$ 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 $D$.