How to prove when this series of polynomials is convergent and when it is divergent?
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Let $P$ and $Q$ be polynomials of degree $k$ and $m$ and suppose $Q(n)ne0forall ninmathbb N$.
Prove that $sum_{n=1}^inftyfrac{P(n)}{Q(n)}$ is convergent if $mge k+2$ and divergent if $mle k+1$.
Can someone point me in the right direction? I have tried to transform into something telescoping and see if the terms cancel each other out but with no result.
Other than that, the only thing I can think of is trying induction but I feel as if there most likely is a much better way to prove it using other thereoms. Possibly the tail lemma.
sequences-and-series polynomials convergence divergent-series
edited Nov 17 at 13:45
Parcly Taxel
41k137198
asked Nov 17 at 13:33
CruZ
153
add a comment |
up vote
0
down vote
favorite
Let $P$ and $Q$ be polynomials of degree $k$ and $m$ and suppose $Q(n)ne0forall ninmathbb N$.
Prove that $sum_{n=1}^inftyfrac{P(n)}{Q(n)}$ is convergent if $mge k+2$ and divergent if $mle k+1$.
Can someone point me in the right direction? I have tried to transform into something telescoping and see if the terms cancel each other out but with no result.
Other than that, the only thing I can think of is trying induction but I feel as if there most likely is a much better way to prove it using other thereoms. Possibly the tail lemma.
sequences-and-series polynomials convergence divergent-series
edited Nov 17 at 13:45
Parcly Taxel
41k137198
asked Nov 17 at 13:33
CruZ
153
Before you start to prove anything you should (I suggest) ask yourself how big/small (roughly speaking) the $n$-th term is. It seems to me that it should be rather like $n^{k-m}$. Oh, maybe I should compare my series with ....
– ancientmathematician
Nov 17 at 14:17
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $P$ and $Q$ be polynomials of degree $k$ and $m$ and suppose $Q(n)ne0forall ninmathbb N$.
Prove that $sum_{n=1}^inftyfrac{P(n)}{Q(n)}$ is convergent if $mge k+2$ and divergent if $mle k+1$.
Can someone point me in the right direction? I have tried to transform into something telescoping and see if the terms cancel each other out but with no result.
Other than that, the only thing I can think of is trying induction but I feel as if there most likely is a much better way to prove it using other thereoms. Possibly the tail lemma.
sequences-and-series polynomials convergence divergent-series
edited Nov 17 at 13:45
Parcly Taxel
41k137198
asked Nov 17 at 13:33
CruZ
153
Let $P$ and $Q$ be polynomials of degree $k$ and $m$ and suppose $Q(n)ne0forall ninmathbb N$.
Prove that $sum_{n=1}^inftyfrac{P(n)}{Q(n)}$ is convergent if $mge k+2$ and divergent if $mle k+1$.
Can someone point me in the right direction? I have tried to transform into something telescoping and see if the terms cancel each other out but with no result.
Other than that, the only thing I can think of is trying induction but I feel as if there most likely is a much better way to prove it using other thereoms. Possibly the tail lemma.
sequences-and-series polynomials convergence divergent-series
sequences-and-series polynomials convergence divergent-series
edited Nov 17 at 13:45
Parcly Taxel
41k137198
asked Nov 17 at 13:33
CruZ
153
edited Nov 17 at 13:45
Parcly Taxel
41k137198
asked Nov 17 at 13:33
CruZ
153
edited Nov 17 at 13:45
Parcly Taxel
41k137198
edited Nov 17 at 13:45
Parcly Taxel
41k137198
edited Nov 17 at 13:45
Parcly Taxel
41k137198
41k137198
asked Nov 17 at 13:33
CruZ
153
asked Nov 17 at 13:33
CruZ
153
asked Nov 17 at 13:33
CruZ
153
153
Before you start to prove anything you should (I suggest) ask yourself how big/small (roughly speaking) the $n$-th term is. It seems to me that it should be rather like $n^{k-m}$. Oh, maybe I should compare my series with ....
– ancientmathematician
Nov 17 at 14:17
add a comment |
Before you start to prove anything you should (I suggest) ask yourself how big/small (roughly speaking) the $n$-th term is. It seems to me that it should be rather like $n^{k-m}$. Oh, maybe I should compare my series with ....
– ancientmathematician
Nov 17 at 14:17
Before you start to prove anything you should (I suggest) ask yourself how big/small (roughly speaking) the $n$-th term is. It seems to me that it should be rather like $n^{k-m}$. Oh, maybe I should compare my series with ....
– ancientmathematician
Nov 17 at 14:17
Before you start to prove anything you should (I suggest) ask yourself how big/small (roughly speaking) the $n$-th term is. It seems to me that it should be rather like $n^{k-m}$. Oh, maybe I should compare my series with ....
– ancientmathematician
Nov 17 at 14:17
add a comment |
1 Answer
1
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oldest
votes
up vote
0
down vote
accepted
If $m ge k+2$ and $n$ is large enough, then there exists $c_1>0$ such that
$$left| frac{Q(n)}{P(n)}right| ge c_1n^{m-k} ge c_1n^2$$
that is $$left| frac{P(n)}{Q(n)}right| le frac{c_1}{n^{2}}$$
Hence by comparison test, we have the result.
Also, if $m le k+1$, WLOG, we assume that $P$ and $Q$ have positive leading coefficient. (Suppose not, we study its negative instead).
If $n$ is large enough, there exists $c_2>0$ such that
$$frac{P(n)}{Q(n)} ge c_2n^{k-m} ge frac{c_2}{n}$$
Again, by comparison test, we have the result.
answered Nov 17 at 14:29
Siong Thye Goh
94.9k1462115
Thank you, that helped a lot!
– CruZ
Nov 18 at 8:53
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
If $m ge k+2$ and $n$ is large enough, then there exists $c_1>0$ such that
$$left| frac{Q(n)}{P(n)}right| ge c_1n^{m-k} ge c_1n^2$$
that is $$left| frac{P(n)}{Q(n)}right| le frac{c_1}{n^{2}}$$
Hence by comparison test, we have the result.
Also, if $m le k+1$, WLOG, we assume that $P$ and $Q$ have positive leading coefficient. (Suppose not, we study its negative instead).
If $n$ is large enough, there exists $c_2>0$ such that
$$frac{P(n)}{Q(n)} ge c_2n^{k-m} ge frac{c_2}{n}$$
Again, by comparison test, we have the result.
answered Nov 17 at 14:29
Siong Thye Goh
94.9k1462115
Thank you, that helped a lot!
– CruZ
Nov 18 at 8:53
add a comment |
up vote
0
down vote
accepted
If $m ge k+2$ and $n$ is large enough, then there exists $c_1>0$ such that
$$left| frac{Q(n)}{P(n)}right| ge c_1n^{m-k} ge c_1n^2$$
that is $$left| frac{P(n)}{Q(n)}right| le frac{c_1}{n^{2}}$$
Hence by comparison test, we have the result.
Also, if $m le k+1$, WLOG, we assume that $P$ and $Q$ have positive leading coefficient. (Suppose not, we study its negative instead).
If $n$ is large enough, there exists $c_2>0$ such that
$$frac{P(n)}{Q(n)} ge c_2n^{k-m} ge frac{c_2}{n}$$
Again, by comparison test, we have the result.
answered Nov 17 at 14:29
Siong Thye Goh
94.9k1462115
Thank you, that helped a lot!
– CruZ
Nov 18 at 8:53
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
If $m ge k+2$ and $n$ is large enough, then there exists $c_1>0$ such that
$$left| frac{Q(n)}{P(n)}right| ge c_1n^{m-k} ge c_1n^2$$
that is $$left| frac{P(n)}{Q(n)}right| le frac{c_1}{n^{2}}$$
Hence by comparison test, we have the result.
Also, if $m le k+1$, WLOG, we assume that $P$ and $Q$ have positive leading coefficient. (Suppose not, we study its negative instead).
If $n$ is large enough, there exists $c_2>0$ such that
$$frac{P(n)}{Q(n)} ge c_2n^{k-m} ge frac{c_2}{n}$$
Again, by comparison test, we have the result.
answered Nov 17 at 14:29
Siong Thye Goh
94.9k1462115
If $m ge k+2$ and $n$ is large enough, then there exists $c_1>0$ such that
$$left| frac{Q(n)}{P(n)}right| ge c_1n^{m-k} ge c_1n^2$$
that is $$left| frac{P(n)}{Q(n)}right| le frac{c_1}{n^{2}}$$
Hence by comparison test, we have the result.
Also, if $m le k+1$, WLOG, we assume that $P$ and $Q$ have positive leading coefficient. (Suppose not, we study its negative instead).
If $n$ is large enough, there exists $c_2>0$ such that
$$frac{P(n)}{Q(n)} ge c_2n^{k-m} ge frac{c_2}{n}$$
Again, by comparison test, we have the result.
answered Nov 17 at 14:29
Siong Thye Goh
94.9k1462115
answered Nov 17 at 14:29
Siong Thye Goh
94.9k1462115
answered Nov 17 at 14:29
Siong Thye Goh
94.9k1462115
answered Nov 17 at 14:29
Siong Thye Goh
94.9k1462115
94.9k1462115
Thank you, that helped a lot!
– CruZ
Nov 18 at 8:53
add a comment |
Thank you, that helped a lot!
– CruZ
Nov 18 at 8:53
Thank you, that helped a lot!
– CruZ
Nov 18 at 8:53
Thank you, that helped a lot!
– CruZ
Nov 18 at 8:53
add a comment |
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Before you start to prove anything you should (I suggest) ask yourself how big/small (roughly speaking) the $n$-th term is. It seems to me that it should be rather like $n^{k-m}$. Oh, maybe I should compare my series with ....
– ancientmathematician
Nov 17 at 14:17