Is $sum _{n=1} ^ infty frac{1}{(x+pi)^2 cdot n^2 }$ uniformly convergent on $(-pi , pi)$?












2












$begingroup$



Is this series uniformly convergent on $(-pi , pi)$:
$$sum _{n=1} ^ infty frac{1}{(x+pi)^2 cdot n^2 },?$$




My Attempt:



If the series were convergent we would have got a natural number $k$ for a fixed $epsilon>0$ such that $sum _{n=k} ^ infty frac{1}{(x+pi)^2 cdot n^2 } < epsilon$ for all $x in (-pi , pi)$. But for this case we will get the term $f_n$ in the summation greater than $1$ for $-pi + 1/n$.
So for every $n$ we will get a $xin (-pi , pi) $ such that summation at $x$ is greater than $1$.



That's why the series is not uniformly convergent on $(-pi , pi)$.



Can somebody please tell me if I have gone wrong anywhere?






real-analysis sequences-and-series analysis convergence uniform-convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Edited. Now is it okay? @Winther
    $endgroup$
    – cmi
    Dec 14 '18 at 20:36










  • $begingroup$
    Do you mean uniformly continuous?
    $endgroup$
    – Mostafa Ayaz
    Dec 14 '18 at 20:37










  • $begingroup$
    no convergent..@MostafaAyaz
    $endgroup$
    – cmi
    Dec 14 '18 at 20:40
















2












$begingroup$



Is this series uniformly convergent on $(-pi , pi)$:
$$sum _{n=1} ^ infty frac{1}{(x+pi)^2 cdot n^2 },?$$




My Attempt:



If the series were convergent we would have got a natural number $k$ for a fixed $epsilon>0$ such that $sum _{n=k} ^ infty frac{1}{(x+pi)^2 cdot n^2 } < epsilon$ for all $x in (-pi , pi)$. But for this case we will get the term $f_n$ in the summation greater than $1$ for $-pi + 1/n$.
So for every $n$ we will get a $xin (-pi , pi) $ such that summation at $x$ is greater than $1$.



That's why the series is not uniformly convergent on $(-pi , pi)$.



Can somebody please tell me if I have gone wrong anywhere?






real-analysis sequences-and-series analysis convergence uniform-convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Edited. Now is it okay? @Winther
    $endgroup$
    – cmi
    Dec 14 '18 at 20:36










  • $begingroup$
    Do you mean uniformly continuous?
    $endgroup$
    – Mostafa Ayaz
    Dec 14 '18 at 20:37










  • $begingroup$
    no convergent..@MostafaAyaz
    $endgroup$
    – cmi
    Dec 14 '18 at 20:40














2












2








2





$begingroup$



Is this series uniformly convergent on $(-pi , pi)$:
$$sum _{n=1} ^ infty frac{1}{(x+pi)^2 cdot n^2 },?$$




My Attempt:



If the series were convergent we would have got a natural number $k$ for a fixed $epsilon>0$ such that $sum _{n=k} ^ infty frac{1}{(x+pi)^2 cdot n^2 } < epsilon$ for all $x in (-pi , pi)$. But for this case we will get the term $f_n$ in the summation greater than $1$ for $-pi + 1/n$.
So for every $n$ we will get a $xin (-pi , pi) $ such that summation at $x$ is greater than $1$.



That's why the series is not uniformly convergent on $(-pi , pi)$.



Can somebody please tell me if I have gone wrong anywhere?






real-analysis sequences-and-series analysis convergence uniform-convergence







share|cite|improve this question











$endgroup$





Is this series uniformly convergent on $(-pi , pi)$:
$$sum _{n=1} ^ infty frac{1}{(x+pi)^2 cdot n^2 },?$$




My Attempt:



If the series were convergent we would have got a natural number $k$ for a fixed $epsilon>0$ such that $sum _{n=k} ^ infty frac{1}{(x+pi)^2 cdot n^2 } < epsilon$ for all $x in (-pi , pi)$. But for this case we will get the term $f_n$ in the summation greater than $1$ for $-pi + 1/n$.
So for every $n$ we will get a $xin (-pi , pi) $ such that summation at $x$ is greater than $1$.



That's why the series is not uniformly convergent on $(-pi , pi)$.



Can somebody please tell me if I have gone wrong anywhere?






real-analysis sequences-and-series analysis convergence uniform-convergence




real-analysis sequences-and-series analysis convergence uniform-convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 21:00









Batominovski

33.1k33293




33.1k33293










asked Dec 14 '18 at 20:27









cmicmi

1,120312




1,120312












  • $begingroup$
    Edited. Now is it okay? @Winther
    $endgroup$
    – cmi
    Dec 14 '18 at 20:36










  • $begingroup$
    Do you mean uniformly continuous?
    $endgroup$
    – Mostafa Ayaz
    Dec 14 '18 at 20:37










  • $begingroup$
    no convergent..@MostafaAyaz
    $endgroup$
    – cmi
    Dec 14 '18 at 20:40


















  • $begingroup$
    Edited. Now is it okay? @Winther
    $endgroup$
    – cmi
    Dec 14 '18 at 20:36










  • $begingroup$
    Do you mean uniformly continuous?
    $endgroup$
    – Mostafa Ayaz
    Dec 14 '18 at 20:37










  • $begingroup$
    no convergent..@MostafaAyaz
    $endgroup$
    – cmi
    Dec 14 '18 at 20:40
















$begingroup$
Edited. Now is it okay? @Winther
$endgroup$
– cmi
Dec 14 '18 at 20:36




$begingroup$
Edited. Now is it okay? @Winther
$endgroup$
– cmi
Dec 14 '18 at 20:36












$begingroup$
Do you mean uniformly continuous?
$endgroup$
– Mostafa Ayaz
Dec 14 '18 at 20:37




$begingroup$
Do you mean uniformly continuous?
$endgroup$
– Mostafa Ayaz
Dec 14 '18 at 20:37












$begingroup$
no convergent..@MostafaAyaz
$endgroup$
– cmi
Dec 14 '18 at 20:40




$begingroup$
no convergent..@MostafaAyaz
$endgroup$
– cmi
Dec 14 '18 at 20:40










2 Answers
2






active

oldest

votes


















2












$begingroup$

If a series uniformly converges, then its summands must approach zero uniformly.



Clearly, the summands do not converge uniformly to $0$.



To see this, take $epsilon=1$.



Then, for all $N$, take any $n>N$ and $x=-pi +frac1n$ and find that



$$left|frac{1}{left(x+piright)^2 n^2}right|=left|frac{1}{left(-pi+frac1n+piright)^2 n^2}right|=1$$



This negates the uniform convergence of the summand $frac{1}{(x+pi)^2n^2}$. And we are done!




If one does not wish to rely on the aforementioned requirement for a series to converge uniformly, then we can proceed brute force as follows.



With $epsilon=1$, we have for all $N>1$



$$begin{align}
left|sum_{n=1}^infty frac{1}{(x+pi)^2n^2}-sum_{n=1}^N frac{1}{(x+pi)^2n^2} right|&=sum_{n=N+1}^infty frac{1}{(x+pi)^2n^2}\\
&ge frac1{(x+pi)^2(N+1)^2}\\
&ge1
end{align}$$



whenever $x= -pi+frac1{ N+1}$. And we are done!






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your solution is fine. Can you please check mine? @Mark Viola
    $endgroup$
    – cmi
    Dec 15 '18 at 4:32










  • $begingroup$
    Your solution is flawed. The summation index is a dummy index. So, your saying $x=-pi +1/n$ makes no sense.
    $endgroup$
    – Mark Viola
    Dec 15 '18 at 14:59



















2












$begingroup$

Your answer is correct. The series isn't uniformly convergent.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The OP's reasoning is flawed. Hence, your statement "Your answer is correct," is misleading, at best.
    $endgroup$
    – Mark Viola
    Dec 16 '18 at 15:37











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

If a series uniformly converges, then its summands must approach zero uniformly.



Clearly, the summands do not converge uniformly to $0$.



To see this, take $epsilon=1$.



Then, for all $N$, take any $n>N$ and $x=-pi +frac1n$ and find that



$$left|frac{1}{left(x+piright)^2 n^2}right|=left|frac{1}{left(-pi+frac1n+piright)^2 n^2}right|=1$$



This negates the uniform convergence of the summand $frac{1}{(x+pi)^2n^2}$. And we are done!




If one does not wish to rely on the aforementioned requirement for a series to converge uniformly, then we can proceed brute force as follows.



With $epsilon=1$, we have for all $N>1$



$$begin{align}
left|sum_{n=1}^infty frac{1}{(x+pi)^2n^2}-sum_{n=1}^N frac{1}{(x+pi)^2n^2} right|&=sum_{n=N+1}^infty frac{1}{(x+pi)^2n^2}\\
&ge frac1{(x+pi)^2(N+1)^2}\\
&ge1
end{align}$$



whenever $x= -pi+frac1{ N+1}$. And we are done!






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your solution is fine. Can you please check mine? @Mark Viola
    $endgroup$
    – cmi
    Dec 15 '18 at 4:32










  • $begingroup$
    Your solution is flawed. The summation index is a dummy index. So, your saying $x=-pi +1/n$ makes no sense.
    $endgroup$
    – Mark Viola
    Dec 15 '18 at 14:59
















2












$begingroup$

If a series uniformly converges, then its summands must approach zero uniformly.



Clearly, the summands do not converge uniformly to $0$.



To see this, take $epsilon=1$.



Then, for all $N$, take any $n>N$ and $x=-pi +frac1n$ and find that



$$left|frac{1}{left(x+piright)^2 n^2}right|=left|frac{1}{left(-pi+frac1n+piright)^2 n^2}right|=1$$



This negates the uniform convergence of the summand $frac{1}{(x+pi)^2n^2}$. And we are done!




If one does not wish to rely on the aforementioned requirement for a series to converge uniformly, then we can proceed brute force as follows.



With $epsilon=1$, we have for all $N>1$



$$begin{align}
left|sum_{n=1}^infty frac{1}{(x+pi)^2n^2}-sum_{n=1}^N frac{1}{(x+pi)^2n^2} right|&=sum_{n=N+1}^infty frac{1}{(x+pi)^2n^2}\\
&ge frac1{(x+pi)^2(N+1)^2}\\
&ge1
end{align}$$



whenever $x= -pi+frac1{ N+1}$. And we are done!






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your solution is fine. Can you please check mine? @Mark Viola
    $endgroup$
    – cmi
    Dec 15 '18 at 4:32










  • $begingroup$
    Your solution is flawed. The summation index is a dummy index. So, your saying $x=-pi +1/n$ makes no sense.
    $endgroup$
    – Mark Viola
    Dec 15 '18 at 14:59














2












2








2





$begingroup$

If a series uniformly converges, then its summands must approach zero uniformly.



Clearly, the summands do not converge uniformly to $0$.



To see this, take $epsilon=1$.



Then, for all $N$, take any $n>N$ and $x=-pi +frac1n$ and find that



$$left|frac{1}{left(x+piright)^2 n^2}right|=left|frac{1}{left(-pi+frac1n+piright)^2 n^2}right|=1$$



This negates the uniform convergence of the summand $frac{1}{(x+pi)^2n^2}$. And we are done!




If one does not wish to rely on the aforementioned requirement for a series to converge uniformly, then we can proceed brute force as follows.



With $epsilon=1$, we have for all $N>1$



$$begin{align}
left|sum_{n=1}^infty frac{1}{(x+pi)^2n^2}-sum_{n=1}^N frac{1}{(x+pi)^2n^2} right|&=sum_{n=N+1}^infty frac{1}{(x+pi)^2n^2}\\
&ge frac1{(x+pi)^2(N+1)^2}\\
&ge1
end{align}$$



whenever $x= -pi+frac1{ N+1}$. And we are done!






share|cite|improve this answer











$endgroup$



If a series uniformly converges, then its summands must approach zero uniformly.



Clearly, the summands do not converge uniformly to $0$.



To see this, take $epsilon=1$.



Then, for all $N$, take any $n>N$ and $x=-pi +frac1n$ and find that



$$left|frac{1}{left(x+piright)^2 n^2}right|=left|frac{1}{left(-pi+frac1n+piright)^2 n^2}right|=1$$



This negates the uniform convergence of the summand $frac{1}{(x+pi)^2n^2}$. And we are done!




If one does not wish to rely on the aforementioned requirement for a series to converge uniformly, then we can proceed brute force as follows.



With $epsilon=1$, we have for all $N>1$



$$begin{align}
left|sum_{n=1}^infty frac{1}{(x+pi)^2n^2}-sum_{n=1}^N frac{1}{(x+pi)^2n^2} right|&=sum_{n=N+1}^infty frac{1}{(x+pi)^2n^2}\\
&ge frac1{(x+pi)^2(N+1)^2}\\
&ge1
end{align}$$



whenever $x= -pi+frac1{ N+1}$. And we are done!







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 14 '18 at 21:31

























answered Dec 14 '18 at 21:10









Mark ViolaMark Viola

133k1277175




133k1277175












  • $begingroup$
    Your solution is fine. Can you please check mine? @Mark Viola
    $endgroup$
    – cmi
    Dec 15 '18 at 4:32










  • $begingroup$
    Your solution is flawed. The summation index is a dummy index. So, your saying $x=-pi +1/n$ makes no sense.
    $endgroup$
    – Mark Viola
    Dec 15 '18 at 14:59


















  • $begingroup$
    Your solution is fine. Can you please check mine? @Mark Viola
    $endgroup$
    – cmi
    Dec 15 '18 at 4:32










  • $begingroup$
    Your solution is flawed. The summation index is a dummy index. So, your saying $x=-pi +1/n$ makes no sense.
    $endgroup$
    – Mark Viola
    Dec 15 '18 at 14:59
















$begingroup$
Your solution is fine. Can you please check mine? @Mark Viola
$endgroup$
– cmi
Dec 15 '18 at 4:32




$begingroup$
Your solution is fine. Can you please check mine? @Mark Viola
$endgroup$
– cmi
Dec 15 '18 at 4:32












$begingroup$
Your solution is flawed. The summation index is a dummy index. So, your saying $x=-pi +1/n$ makes no sense.
$endgroup$
– Mark Viola
Dec 15 '18 at 14:59




$begingroup$
Your solution is flawed. The summation index is a dummy index. So, your saying $x=-pi +1/n$ makes no sense.
$endgroup$
– Mark Viola
Dec 15 '18 at 14:59











2












$begingroup$

Your answer is correct. The series isn't uniformly convergent.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The OP's reasoning is flawed. Hence, your statement "Your answer is correct," is misleading, at best.
    $endgroup$
    – Mark Viola
    Dec 16 '18 at 15:37
















2












$begingroup$

Your answer is correct. The series isn't uniformly convergent.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The OP's reasoning is flawed. Hence, your statement "Your answer is correct," is misleading, at best.
    $endgroup$
    – Mark Viola
    Dec 16 '18 at 15:37














2












2








2





$begingroup$

Your answer is correct. The series isn't uniformly convergent.






share|cite|improve this answer











$endgroup$



Your answer is correct. The series isn't uniformly convergent.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 14 '18 at 22:07









Jam

4,98921431




4,98921431










answered Dec 14 '18 at 20:43









moutheticsmouthetics

51037




51037












  • $begingroup$
    The OP's reasoning is flawed. Hence, your statement "Your answer is correct," is misleading, at best.
    $endgroup$
    – Mark Viola
    Dec 16 '18 at 15:37


















  • $begingroup$
    The OP's reasoning is flawed. Hence, your statement "Your answer is correct," is misleading, at best.
    $endgroup$
    – Mark Viola
    Dec 16 '18 at 15:37
















$begingroup$
The OP's reasoning is flawed. Hence, your statement "Your answer is correct," is misleading, at best.
$endgroup$
– Mark Viola
Dec 16 '18 at 15:37




$begingroup$
The OP's reasoning is flawed. Hence, your statement "Your answer is correct," is misleading, at best.
$endgroup$
– Mark Viola
Dec 16 '18 at 15:37


















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