Mixing
Series and polynomials
$begingroup$
Suppose $p$ is a polynomial of degree $geq 2$ such that any integer is not a root of $p$, i.e, $p(n)neq 0;forall,ninmathbb{Z}$.
Is it true that
$$sum_{n=-infty}^{infty}frac{1}{p(n)}<infty$$
for any polynomial $p$ satisfying the above conditions and with real coefficients? If not, what about
$$sum_{n=0}^{infty}frac{1}{p(n)}?$$
sequences-and-series polynomials convergence
asked Feb 27 '16 at 5:04
LararaLarara
827519
$endgroup$
add a comment |
$begingroup$
Suppose $p$ is a polynomial of degree $geq 2$ such that any integer is not a root of $p$, i.e, $p(n)neq 0;forall,ninmathbb{Z}$.
Is it true that
$$sum_{n=-infty}^{infty}frac{1}{p(n)}<infty$$
for any polynomial $p$ satisfying the above conditions and with real coefficients? If not, what about
$$sum_{n=0}^{infty}frac{1}{p(n)}?$$
sequences-and-series polynomials convergence
asked Feb 27 '16 at 5:04
LararaLarara
827519
$endgroup$
$begingroup$
I have written an Edit to my answer making the link with complex function theory.
$endgroup$
– Jean Marie
Jan 9 at 10:07
add a comment |
$begingroup$
Suppose $p$ is a polynomial of degree $geq 2$ such that any integer is not a root of $p$, i.e, $p(n)neq 0;forall,ninmathbb{Z}$.
Is it true that
$$sum_{n=-infty}^{infty}frac{1}{p(n)}<infty$$
for any polynomial $p$ satisfying the above conditions and with real coefficients? If not, what about
$$sum_{n=0}^{infty}frac{1}{p(n)}?$$
sequences-and-series polynomials convergence
asked Feb 27 '16 at 5:04
LararaLarara
827519
$endgroup$
Suppose $p$ is a polynomial of degree $geq 2$ such that any integer is not a root of $p$, i.e, $p(n)neq 0;forall,ninmathbb{Z}$.
Is it true that
$$sum_{n=-infty}^{infty}frac{1}{p(n)}<infty$$
for any polynomial $p$ satisfying the above conditions and with real coefficients? If not, what about
$$sum_{n=0}^{infty}frac{1}{p(n)}?$$
sequences-and-series polynomials convergence
sequences-and-series polynomials convergence
asked Feb 27 '16 at 5:04
LararaLarara
827519
asked Feb 27 '16 at 5:04
LararaLarara
827519
asked Feb 27 '16 at 5:04
LararaLarara
827519
asked Feb 27 '16 at 5:04
LararaLarara
827519
asked Feb 27 '16 at 5:04
LararaLarara
827519
827519
$begingroup$
I have written an Edit to my answer making the link with complex function theory.
$endgroup$
– Jean Marie
Jan 9 at 10:07
add a comment |
$begingroup$
I have written an Edit to my answer making the link with complex function theory.
$endgroup$
– Jean Marie
Jan 9 at 10:07
$begingroup$
I have written an Edit to my answer making the link with complex function theory.
$endgroup$
– Jean Marie
Jan 9 at 10:07
$begingroup$
I have written an Edit to my answer making the link with complex function theory.
$endgroup$
– Jean Marie
Jan 9 at 10:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, they both converge. If the polynomial is of degree $geq 2$ then, for $|n|$ large enough, we have
$$
-frac{1}{n^2}leqfrac{1}{p(n)}leq frac{1}{n^2}.
$$
Since $sum_{n=0}^infty frac{1}{n^2}$ converges, then also $sum_{n=0}^infty frac{1}{p(n)}$ must converge (absolutely) by the comparison test (and same for the series over $mathbb{Z}$).
answered Feb 27 '16 at 5:15
bartgolbartgol
5,0321419
$endgroup$
2
$begingroup$
No, I am sorry, you have not $-frac{pi^2}{6}leqsum_{n=0}^{infty}frac{1}{p(n)}leqfrac{pi^2}{6}$ because in the proof, you have to drop a certain number of terms. Another even simpler reason explaining that the series $sum_{n=0}^{infty}frac{1}{p(n)}$ can take any real value is that you can multiply one of its values by any real you want.
$endgroup$
– Jean Marie
Feb 27 '16 at 5:27
add a comment |
$begingroup$
Yes, there is always convergence.
Consider the second degree case (the same demonstration could be given for any degree $>2$).
We can restrict our attention as well to the part with positive indices:
$$sum_{n=0}^{infty}frac{1}{p(n)}=sum_{n=0}^{infty}frac{1}{an^2+bn+c}$$
(the other part can be treated in the same way).
We may also assume that $a>0$ (otherwise we take the opposite of the series).
As
$$lim_{n to infty} frac{an^2+bn+c}{(a/2)n^2} = 2$$
we can say that, there exists a $n_0$ such that :
$$text{for any} n>n_0, frac{an^2+bn+c}{(a/2)n^2} > 1$$
i.e.,
$$text{for} n>n_0, an^2+bn+c > frac{a}{2}n^2 > 0$$
Thus
$$0<sum_{n=n_0+1}^{infty}frac{1}{an^2+bn+c}<frac{2}{a}sum_{n=n_0+1}^{infty}frac{1}{n^2}$$
which is known to be convergent.
The nature of a series (convergent / not convergent) being not entailed by the fact that a finite number of terms are dropped in it, our proof is completed.
Remark : using complex function theory, more precisely residue theorem, under the condition that $p(x)$ has simple roots, one can express the second series as a certain sum.
$$sum_{-infty}^{infty}frac{1}{p(n)}=-text{sum of values of} dfrac{pi cot(pi x)}{p(x)} text{at roots of} p.$$
(see http://www.supermath.info/InfiniteSeriesandtheResidueTheorem.pdf)
For example, if $p(x)=x^2+x+1$,
$$sum_{-infty}^{infty}frac{1}{p(n)}=-r(omega)-r(overline{omega})= dfrac{2 pi}{sqrt{3}} tanh(pi dfrac{sqrt{3}}{2}),$$
where $omega=e^{2ipi/3}$ and $overline{omega}$ are the roots of $p(x)=0$ and $r(x):=dfrac{pi cot(pi x)}{p'(x)}$ (classical formula for the residue at a simple pole).
answered Feb 27 '16 at 5:21
Jean MarieJean Marie
29.5k42050
$endgroup$
$begingroup$
If $a>0$ then don't we have, for $n$ large enough, $an^2+bn+c>0$?
$endgroup$
– Larara
Feb 27 '16 at 5:26
$begingroup$
Yes, that's true.
$endgroup$
– Jean Marie
Feb 27 '16 at 5:28
$begingroup$
But the first inequality in your answer says the contrary. Also, I did not understand from where the $frac{1}{2}an^2$ comes from.
$endgroup$
– Larara
Feb 27 '16 at 5:30
$begingroup$
I have corrected an error in the last formula (2/a instead of 1/(2a)). Is that the problem you had ?
$endgroup$
– Jean Marie
Feb 27 '16 at 5:35
$begingroup$
Do you agree that if the sum of your series is $S$, you only have to divide coefficients $a,b,c$ by 2 in order to get a sum of 2S, whatever the value of $S$ ?
$endgroup$
– Jean Marie
Feb 27 '16 at 5:38
|
show 1 more comment
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, they both converge. If the polynomial is of degree $geq 2$ then, for $|n|$ large enough, we have
$$
-frac{1}{n^2}leqfrac{1}{p(n)}leq frac{1}{n^2}.
$$
Since $sum_{n=0}^infty frac{1}{n^2}$ converges, then also $sum_{n=0}^infty frac{1}{p(n)}$ must converge (absolutely) by the comparison test (and same for the series over $mathbb{Z}$).
answered Feb 27 '16 at 5:15
bartgolbartgol
5,0321419
$endgroup$
2
$begingroup$
No, I am sorry, you have not $-frac{pi^2}{6}leqsum_{n=0}^{infty}frac{1}{p(n)}leqfrac{pi^2}{6}$ because in the proof, you have to drop a certain number of terms. Another even simpler reason explaining that the series $sum_{n=0}^{infty}frac{1}{p(n)}$ can take any real value is that you can multiply one of its values by any real you want.
$endgroup$
– Jean Marie
Feb 27 '16 at 5:27
add a comment |
$begingroup$
Yes, they both converge. If the polynomial is of degree $geq 2$ then, for $|n|$ large enough, we have
$$
-frac{1}{n^2}leqfrac{1}{p(n)}leq frac{1}{n^2}.
$$
Since $sum_{n=0}^infty frac{1}{n^2}$ converges, then also $sum_{n=0}^infty frac{1}{p(n)}$ must converge (absolutely) by the comparison test (and same for the series over $mathbb{Z}$).
answered Feb 27 '16 at 5:15
bartgolbartgol
5,0321419
$endgroup$
2
$begingroup$
No, I am sorry, you have not $-frac{pi^2}{6}leqsum_{n=0}^{infty}frac{1}{p(n)}leqfrac{pi^2}{6}$ because in the proof, you have to drop a certain number of terms. Another even simpler reason explaining that the series $sum_{n=0}^{infty}frac{1}{p(n)}$ can take any real value is that you can multiply one of its values by any real you want.
$endgroup$
– Jean Marie
Feb 27 '16 at 5:27
add a comment |
$begingroup$
Yes, they both converge. If the polynomial is of degree $geq 2$ then, for $|n|$ large enough, we have
$$
-frac{1}{n^2}leqfrac{1}{p(n)}leq frac{1}{n^2}.
$$
Since $sum_{n=0}^infty frac{1}{n^2}$ converges, then also $sum_{n=0}^infty frac{1}{p(n)}$ must converge (absolutely) by the comparison test (and same for the series over $mathbb{Z}$).
answered Feb 27 '16 at 5:15
bartgolbartgol
5,0321419
$endgroup$
Yes, they both converge. If the polynomial is of degree $geq 2$ then, for $|n|$ large enough, we have
$$
-frac{1}{n^2}leqfrac{1}{p(n)}leq frac{1}{n^2}.
$$
Since $sum_{n=0}^infty frac{1}{n^2}$ converges, then also $sum_{n=0}^infty frac{1}{p(n)}$ must converge (absolutely) by the comparison test (and same for the series over $mathbb{Z}$).
answered Feb 27 '16 at 5:15
bartgolbartgol
5,0321419
answered Feb 27 '16 at 5:15
bartgolbartgol
5,0321419
answered Feb 27 '16 at 5:15
bartgolbartgol
5,0321419
answered Feb 27 '16 at 5:15
bartgolbartgol
5,0321419
5,0321419
2
$begingroup$
No, I am sorry, you have not $-frac{pi^2}{6}leqsum_{n=0}^{infty}frac{1}{p(n)}leqfrac{pi^2}{6}$ because in the proof, you have to drop a certain number of terms. Another even simpler reason explaining that the series $sum_{n=0}^{infty}frac{1}{p(n)}$ can take any real value is that you can multiply one of its values by any real you want.
$endgroup$
– Jean Marie
Feb 27 '16 at 5:27
add a comment |
2
$begingroup$
No, I am sorry, you have not $-frac{pi^2}{6}leqsum_{n=0}^{infty}frac{1}{p(n)}leqfrac{pi^2}{6}$ because in the proof, you have to drop a certain number of terms. Another even simpler reason explaining that the series $sum_{n=0}^{infty}frac{1}{p(n)}$ can take any real value is that you can multiply one of its values by any real you want.
$endgroup$
– Jean Marie
Feb 27 '16 at 5:27
2
2
$begingroup$
No, I am sorry, you have not $-frac{pi^2}{6}leqsum_{n=0}^{infty}frac{1}{p(n)}leqfrac{pi^2}{6}$ because in the proof, you have to drop a certain number of terms. Another even simpler reason explaining that the series $sum_{n=0}^{infty}frac{1}{p(n)}$ can take any real value is that you can multiply one of its values by any real you want.
$endgroup$
– Jean Marie
Feb 27 '16 at 5:27
$begingroup$
No, I am sorry, you have not $-frac{pi^2}{6}leqsum_{n=0}^{infty}frac{1}{p(n)}leqfrac{pi^2}{6}$ because in the proof, you have to drop a certain number of terms. Another even simpler reason explaining that the series $sum_{n=0}^{infty}frac{1}{p(n)}$ can take any real value is that you can multiply one of its values by any real you want.
$endgroup$
– Jean Marie
Feb 27 '16 at 5:27
add a comment |
$begingroup$
Yes, there is always convergence.
Consider the second degree case (the same demonstration could be given for any degree $>2$).
We can restrict our attention as well to the part with positive indices:
$$sum_{n=0}^{infty}frac{1}{p(n)}=sum_{n=0}^{infty}frac{1}{an^2+bn+c}$$
(the other part can be treated in the same way).
We may also assume that $a>0$ (otherwise we take the opposite of the series).
As
$$lim_{n to infty} frac{an^2+bn+c}{(a/2)n^2} = 2$$
we can say that, there exists a $n_0$ such that :
$$text{for any} n>n_0, frac{an^2+bn+c}{(a/2)n^2} > 1$$
i.e.,
$$text{for} n>n_0, an^2+bn+c > frac{a}{2}n^2 > 0$$
Thus
$$0<sum_{n=n_0+1}^{infty}frac{1}{an^2+bn+c}<frac{2}{a}sum_{n=n_0+1}^{infty}frac{1}{n^2}$$
which is known to be convergent.
The nature of a series (convergent / not convergent) being not entailed by the fact that a finite number of terms are dropped in it, our proof is completed.
Remark : using complex function theory, more precisely residue theorem, under the condition that $p(x)$ has simple roots, one can express the second series as a certain sum.
$$sum_{-infty}^{infty}frac{1}{p(n)}=-text{sum of values of} dfrac{pi cot(pi x)}{p(x)} text{at roots of} p.$$
(see http://www.supermath.info/InfiniteSeriesandtheResidueTheorem.pdf)
For example, if $p(x)=x^2+x+1$,
$$sum_{-infty}^{infty}frac{1}{p(n)}=-r(omega)-r(overline{omega})= dfrac{2 pi}{sqrt{3}} tanh(pi dfrac{sqrt{3}}{2}),$$
where $omega=e^{2ipi/3}$ and $overline{omega}$ are the roots of $p(x)=0$ and $r(x):=dfrac{pi cot(pi x)}{p'(x)}$ (classical formula for the residue at a simple pole).
answered Feb 27 '16 at 5:21
Jean MarieJean Marie
29.5k42050
$endgroup$
$begingroup$
If $a>0$ then don't we have, for $n$ large enough, $an^2+bn+c>0$?
$endgroup$
– Larara
Feb 27 '16 at 5:26
$begingroup$
Yes, that's true.
$endgroup$
– Jean Marie
Feb 27 '16 at 5:28
$begingroup$
But the first inequality in your answer says the contrary. Also, I did not understand from where the $frac{1}{2}an^2$ comes from.
$endgroup$
– Larara
Feb 27 '16 at 5:30
$begingroup$
I have corrected an error in the last formula (2/a instead of 1/(2a)). Is that the problem you had ?
$endgroup$
– Jean Marie
Feb 27 '16 at 5:35
$begingroup$
Do you agree that if the sum of your series is $S$, you only have to divide coefficients $a,b,c$ by 2 in order to get a sum of 2S, whatever the value of $S$ ?
$endgroup$
– Jean Marie
Feb 27 '16 at 5:38
|
show 1 more comment
$begingroup$
Yes, there is always convergence.
Consider the second degree case (the same demonstration could be given for any degree $>2$).
We can restrict our attention as well to the part with positive indices:
$$sum_{n=0}^{infty}frac{1}{p(n)}=sum_{n=0}^{infty}frac{1}{an^2+bn+c}$$
(the other part can be treated in the same way).
We may also assume that $a>0$ (otherwise we take the opposite of the series).
As
$$lim_{n to infty} frac{an^2+bn+c}{(a/2)n^2} = 2$$
we can say that, there exists a $n_0$ such that :
$$text{for any} n>n_0, frac{an^2+bn+c}{(a/2)n^2} > 1$$
i.e.,
$$text{for} n>n_0, an^2+bn+c > frac{a}{2}n^2 > 0$$
Thus
$$0<sum_{n=n_0+1}^{infty}frac{1}{an^2+bn+c}<frac{2}{a}sum_{n=n_0+1}^{infty}frac{1}{n^2}$$
which is known to be convergent.
The nature of a series (convergent / not convergent) being not entailed by the fact that a finite number of terms are dropped in it, our proof is completed.
Remark : using complex function theory, more precisely residue theorem, under the condition that $p(x)$ has simple roots, one can express the second series as a certain sum.
$$sum_{-infty}^{infty}frac{1}{p(n)}=-text{sum of values of} dfrac{pi cot(pi x)}{p(x)} text{at roots of} p.$$
(see http://www.supermath.info/InfiniteSeriesandtheResidueTheorem.pdf)
For example, if $p(x)=x^2+x+1$,
$$sum_{-infty}^{infty}frac{1}{p(n)}=-r(omega)-r(overline{omega})= dfrac{2 pi}{sqrt{3}} tanh(pi dfrac{sqrt{3}}{2}),$$
where $omega=e^{2ipi/3}$ and $overline{omega}$ are the roots of $p(x)=0$ and $r(x):=dfrac{pi cot(pi x)}{p'(x)}$ (classical formula for the residue at a simple pole).
answered Feb 27 '16 at 5:21
Jean MarieJean Marie
29.5k42050
$endgroup$
$begingroup$
If $a>0$ then don't we have, for $n$ large enough, $an^2+bn+c>0$?
$endgroup$
– Larara
Feb 27 '16 at 5:26
$begingroup$
Yes, that's true.
$endgroup$
– Jean Marie
Feb 27 '16 at 5:28
$begingroup$
But the first inequality in your answer says the contrary. Also, I did not understand from where the $frac{1}{2}an^2$ comes from.
$endgroup$
– Larara
Feb 27 '16 at 5:30
$begingroup$
I have corrected an error in the last formula (2/a instead of 1/(2a)). Is that the problem you had ?
$endgroup$
– Jean Marie
Feb 27 '16 at 5:35
$begingroup$
Do you agree that if the sum of your series is $S$, you only have to divide coefficients $a,b,c$ by 2 in order to get a sum of 2S, whatever the value of $S$ ?
$endgroup$
– Jean Marie
Feb 27 '16 at 5:38
|
show 1 more comment
$begingroup$
Yes, there is always convergence.
Consider the second degree case (the same demonstration could be given for any degree $>2$).
We can restrict our attention as well to the part with positive indices:
$$sum_{n=0}^{infty}frac{1}{p(n)}=sum_{n=0}^{infty}frac{1}{an^2+bn+c}$$
(the other part can be treated in the same way).
We may also assume that $a>0$ (otherwise we take the opposite of the series).
As
$$lim_{n to infty} frac{an^2+bn+c}{(a/2)n^2} = 2$$
we can say that, there exists a $n_0$ such that :
$$text{for any} n>n_0, frac{an^2+bn+c}{(a/2)n^2} > 1$$
i.e.,
$$text{for} n>n_0, an^2+bn+c > frac{a}{2}n^2 > 0$$
Thus
$$0<sum_{n=n_0+1}^{infty}frac{1}{an^2+bn+c}<frac{2}{a}sum_{n=n_0+1}^{infty}frac{1}{n^2}$$
which is known to be convergent.
The nature of a series (convergent / not convergent) being not entailed by the fact that a finite number of terms are dropped in it, our proof is completed.
Remark : using complex function theory, more precisely residue theorem, under the condition that $p(x)$ has simple roots, one can express the second series as a certain sum.
$$sum_{-infty}^{infty}frac{1}{p(n)}=-text{sum of values of} dfrac{pi cot(pi x)}{p(x)} text{at roots of} p.$$
(see http://www.supermath.info/InfiniteSeriesandtheResidueTheorem.pdf)
For example, if $p(x)=x^2+x+1$,
$$sum_{-infty}^{infty}frac{1}{p(n)}=-r(omega)-r(overline{omega})= dfrac{2 pi}{sqrt{3}} tanh(pi dfrac{sqrt{3}}{2}),$$
where $omega=e^{2ipi/3}$ and $overline{omega}$ are the roots of $p(x)=0$ and $r(x):=dfrac{pi cot(pi x)}{p'(x)}$ (classical formula for the residue at a simple pole).
answered Feb 27 '16 at 5:21
Jean MarieJean Marie
29.5k42050
$endgroup$
Yes, there is always convergence.
Consider the second degree case (the same demonstration could be given for any degree $>2$).
We can restrict our attention as well to the part with positive indices:
$$sum_{n=0}^{infty}frac{1}{p(n)}=sum_{n=0}^{infty}frac{1}{an^2+bn+c}$$
(the other part can be treated in the same way).
We may also assume that $a>0$ (otherwise we take the opposite of the series).
As
$$lim_{n to infty} frac{an^2+bn+c}{(a/2)n^2} = 2$$
we can say that, there exists a $n_0$ such that :
$$text{for any} n>n_0, frac{an^2+bn+c}{(a/2)n^2} > 1$$
i.e.,
$$text{for} n>n_0, an^2+bn+c > frac{a}{2}n^2 > 0$$
Thus
$$0<sum_{n=n_0+1}^{infty}frac{1}{an^2+bn+c}<frac{2}{a}sum_{n=n_0+1}^{infty}frac{1}{n^2}$$
which is known to be convergent.
The nature of a series (convergent / not convergent) being not entailed by the fact that a finite number of terms are dropped in it, our proof is completed.
Remark : using complex function theory, more precisely residue theorem, under the condition that $p(x)$ has simple roots, one can express the second series as a certain sum.
$$sum_{-infty}^{infty}frac{1}{p(n)}=-text{sum of values of} dfrac{pi cot(pi x)}{p(x)} text{at roots of} p.$$
(see http://www.supermath.info/InfiniteSeriesandtheResidueTheorem.pdf)
For example, if $p(x)=x^2+x+1$,
$$sum_{-infty}^{infty}frac{1}{p(n)}=-r(omega)-r(overline{omega})= dfrac{2 pi}{sqrt{3}} tanh(pi dfrac{sqrt{3}}{2}),$$
where $omega=e^{2ipi/3}$ and $overline{omega}$ are the roots of $p(x)=0$ and $r(x):=dfrac{pi cot(pi x)}{p'(x)}$ (classical formula for the residue at a simple pole).
answered Feb 27 '16 at 5:21
Jean MarieJean Marie
29.5k42050
edited Jan 9 at 10:05
answered Feb 27 '16 at 5:21
Jean MarieJean Marie
29.5k42050
answered Feb 27 '16 at 5:21
Jean MarieJean Marie
29.5k42050
answered Feb 27 '16 at 5:21
Jean MarieJean Marie
29.5k42050
29.5k42050
$begingroup$
If $a>0$ then don't we have, for $n$ large enough, $an^2+bn+c>0$?
$endgroup$
– Larara
Feb 27 '16 at 5:26
$begingroup$
Yes, that's true.
$endgroup$
– Jean Marie
Feb 27 '16 at 5:28
$begingroup$
But the first inequality in your answer says the contrary. Also, I did not understand from where the $frac{1}{2}an^2$ comes from.
$endgroup$
– Larara
Feb 27 '16 at 5:30
$begingroup$
I have corrected an error in the last formula (2/a instead of 1/(2a)). Is that the problem you had ?
$endgroup$
– Jean Marie
Feb 27 '16 at 5:35
$begingroup$
Do you agree that if the sum of your series is $S$, you only have to divide coefficients $a,b,c$ by 2 in order to get a sum of 2S, whatever the value of $S$ ?
$endgroup$
– Jean Marie
Feb 27 '16 at 5:38
|
show 1 more comment
$begingroup$
If $a>0$ then don't we have, for $n$ large enough, $an^2+bn+c>0$?
$endgroup$
– Larara
Feb 27 '16 at 5:26
$begingroup$
Yes, that's true.
$endgroup$
– Jean Marie
Feb 27 '16 at 5:28
$begingroup$
But the first inequality in your answer says the contrary. Also, I did not understand from where the $frac{1}{2}an^2$ comes from.
$endgroup$
– Larara
Feb 27 '16 at 5:30
$begingroup$
I have corrected an error in the last formula (2/a instead of 1/(2a)). Is that the problem you had ?
$endgroup$
– Jean Marie
Feb 27 '16 at 5:35
$begingroup$
Do you agree that if the sum of your series is $S$, you only have to divide coefficients $a,b,c$ by 2 in order to get a sum of 2S, whatever the value of $S$ ?
$endgroup$
– Jean Marie
Feb 27 '16 at 5:38
$begingroup$
If $a>0$ then don't we have, for $n$ large enough, $an^2+bn+c>0$?
$endgroup$
– Larara
Feb 27 '16 at 5:26
$begingroup$
If $a>0$ then don't we have, for $n$ large enough, $an^2+bn+c>0$?
$endgroup$
– Larara
Feb 27 '16 at 5:26
$begingroup$
Yes, that's true.
$endgroup$
– Jean Marie
Feb 27 '16 at 5:28
$begingroup$
Yes, that's true.
$endgroup$
– Jean Marie
Feb 27 '16 at 5:28
$begingroup$
But the first inequality in your answer says the contrary. Also, I did not understand from where the $frac{1}{2}an^2$ comes from.
$endgroup$
– Larara
Feb 27 '16 at 5:30
$begingroup$
But the first inequality in your answer says the contrary. Also, I did not understand from where the $frac{1}{2}an^2$ comes from.
$endgroup$
– Larara
Feb 27 '16 at 5:30
$begingroup$
I have corrected an error in the last formula (2/a instead of 1/(2a)). Is that the problem you had ?
$endgroup$
– Jean Marie
Feb 27 '16 at 5:35
$begingroup$
I have corrected an error in the last formula (2/a instead of 1/(2a)). Is that the problem you had ?
$endgroup$
– Jean Marie
Feb 27 '16 at 5:35
$begingroup$
Do you agree that if the sum of your series is $S$, you only have to divide coefficients $a,b,c$ by 2 in order to get a sum of 2S, whatever the value of $S$ ?
$endgroup$
– Jean Marie
Feb 27 '16 at 5:38
$begingroup$
Do you agree that if the sum of your series is $S$, you only have to divide coefficients $a,b,c$ by 2 in order to get a sum of 2S, whatever the value of $S$ ?
$endgroup$
– Jean Marie
Feb 27 '16 at 5:38
|
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$begingroup$
I have written an Edit to my answer making the link with complex function theory.
$endgroup$
– Jean Marie
Jan 9 at 10:07