Prove that the series, $sum_{n=1}^{infty}{x^{2n}over n^2+x^{2n}}$ converges uniformly in $[-1,1]$












2












$begingroup$


Consider the series $$sum_{n=1}^{infty}{x^{2n}over n^2+x^{2n}}$$



I want to show that the series is uniformly convergent in $[-1,1]$.




Theorem: A series of functions $sum f_n$ will converge uniformly on $[a,b]$ if there exist a convergent $sum M_n$ of positive numbers
such that for all $xin [a,b]$ $$|f_n(x)|le M_n$$ for all $n$.




Attempt:



For $xin [-1,1]quad$ $|x^{2n}|le 1tag{1}$.



Recall,




Reverse Triangle Inequality begin{equation*} ||x|-|y||le|x+y|.
end{equation*}




So $n^2-|x^{2n}|le|n^2+x^{2n}|$ Therefore we have $n^2-1le |x^{2n}+n^2|tag{2}$



Using $(1)$ and $(2)$, we have $$|f_n(x)|le {1over n^2-1}$$



Let $a_n=1/n^2$, then ${M_nover a_n}={n^2over n^2-1}={1over 1-{1over n^2}}$ Therefor by Limit form comparison test, $sum M_n$ converges.



The result follows.




Is this series uniformly convergent only in $[-1,1]$ and to what function it converges?



Edits: For $|x|>1$,



Divide by $|x|^{2n}$, we get $${1over 1+{n^2over |x|^{2n}}}$$. Since exponential grows faster than $n^2$ the expression goes to $1$ as $nto infty$



Therefore series diverges for $|x|>1$






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Does the summand go to zero for $x>1$, as $n $ goes to infinity?
    $endgroup$
    – AnyAD
    Oct 23 '18 at 12:05






  • 1




    $begingroup$
    @AnyAD Thanks I understand it now. I have edited the post to show the calculation.
    $endgroup$
    – StammeringMathematician
    Oct 23 '18 at 12:24
















2












$begingroup$


Consider the series $$sum_{n=1}^{infty}{x^{2n}over n^2+x^{2n}}$$



I want to show that the series is uniformly convergent in $[-1,1]$.




Theorem: A series of functions $sum f_n$ will converge uniformly on $[a,b]$ if there exist a convergent $sum M_n$ of positive numbers
such that for all $xin [a,b]$ $$|f_n(x)|le M_n$$ for all $n$.




Attempt:



For $xin [-1,1]quad$ $|x^{2n}|le 1tag{1}$.



Recall,




Reverse Triangle Inequality begin{equation*} ||x|-|y||le|x+y|.
end{equation*}




So $n^2-|x^{2n}|le|n^2+x^{2n}|$ Therefore we have $n^2-1le |x^{2n}+n^2|tag{2}$



Using $(1)$ and $(2)$, we have $$|f_n(x)|le {1over n^2-1}$$



Let $a_n=1/n^2$, then ${M_nover a_n}={n^2over n^2-1}={1over 1-{1over n^2}}$ Therefor by Limit form comparison test, $sum M_n$ converges.



The result follows.




Is this series uniformly convergent only in $[-1,1]$ and to what function it converges?



Edits: For $|x|>1$,



Divide by $|x|^{2n}$, we get $${1over 1+{n^2over |x|^{2n}}}$$. Since exponential grows faster than $n^2$ the expression goes to $1$ as $nto infty$



Therefore series diverges for $|x|>1$






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Does the summand go to zero for $x>1$, as $n $ goes to infinity?
    $endgroup$
    – AnyAD
    Oct 23 '18 at 12:05






  • 1




    $begingroup$
    @AnyAD Thanks I understand it now. I have edited the post to show the calculation.
    $endgroup$
    – StammeringMathematician
    Oct 23 '18 at 12:24














2












2








2





$begingroup$


Consider the series $$sum_{n=1}^{infty}{x^{2n}over n^2+x^{2n}}$$



I want to show that the series is uniformly convergent in $[-1,1]$.




Theorem: A series of functions $sum f_n$ will converge uniformly on $[a,b]$ if there exist a convergent $sum M_n$ of positive numbers
such that for all $xin [a,b]$ $$|f_n(x)|le M_n$$ for all $n$.




Attempt:



For $xin [-1,1]quad$ $|x^{2n}|le 1tag{1}$.



Recall,




Reverse Triangle Inequality begin{equation*} ||x|-|y||le|x+y|.
end{equation*}




So $n^2-|x^{2n}|le|n^2+x^{2n}|$ Therefore we have $n^2-1le |x^{2n}+n^2|tag{2}$



Using $(1)$ and $(2)$, we have $$|f_n(x)|le {1over n^2-1}$$



Let $a_n=1/n^2$, then ${M_nover a_n}={n^2over n^2-1}={1over 1-{1over n^2}}$ Therefor by Limit form comparison test, $sum M_n$ converges.



The result follows.




Is this series uniformly convergent only in $[-1,1]$ and to what function it converges?



Edits: For $|x|>1$,



Divide by $|x|^{2n}$, we get $${1over 1+{n^2over |x|^{2n}}}$$. Since exponential grows faster than $n^2$ the expression goes to $1$ as $nto infty$



Therefore series diverges for $|x|>1$






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$




Consider the series $$sum_{n=1}^{infty}{x^{2n}over n^2+x^{2n}}$$



I want to show that the series is uniformly convergent in $[-1,1]$.




Theorem: A series of functions $sum f_n$ will converge uniformly on $[a,b]$ if there exist a convergent $sum M_n$ of positive numbers
such that for all $xin [a,b]$ $$|f_n(x)|le M_n$$ for all $n$.




Attempt:



For $xin [-1,1]quad$ $|x^{2n}|le 1tag{1}$.



Recall,




Reverse Triangle Inequality begin{equation*} ||x|-|y||le|x+y|.
end{equation*}




So $n^2-|x^{2n}|le|n^2+x^{2n}|$ Therefore we have $n^2-1le |x^{2n}+n^2|tag{2}$



Using $(1)$ and $(2)$, we have $$|f_n(x)|le {1over n^2-1}$$



Let $a_n=1/n^2$, then ${M_nover a_n}={n^2over n^2-1}={1over 1-{1over n^2}}$ Therefor by Limit form comparison test, $sum M_n$ converges.



The result follows.




Is this series uniformly convergent only in $[-1,1]$ and to what function it converges?



Edits: For $|x|>1$,



Divide by $|x|^{2n}$, we get $${1over 1+{n^2over |x|^{2n}}}$$. Since exponential grows faster than $n^2$ the expression goes to $1$ as $nto infty$



Therefore series diverges for $|x|>1$






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 23 '18 at 12:23







StammeringMathematician

















asked Oct 23 '18 at 11:56









StammeringMathematicianStammeringMathematician

2,5541323




2,5541323








  • 1




    $begingroup$
    Does the summand go to zero for $x>1$, as $n $ goes to infinity?
    $endgroup$
    – AnyAD
    Oct 23 '18 at 12:05






  • 1




    $begingroup$
    @AnyAD Thanks I understand it now. I have edited the post to show the calculation.
    $endgroup$
    – StammeringMathematician
    Oct 23 '18 at 12:24














  • 1




    $begingroup$
    Does the summand go to zero for $x>1$, as $n $ goes to infinity?
    $endgroup$
    – AnyAD
    Oct 23 '18 at 12:05






  • 1




    $begingroup$
    @AnyAD Thanks I understand it now. I have edited the post to show the calculation.
    $endgroup$
    – StammeringMathematician
    Oct 23 '18 at 12:24








1




1




$begingroup$
Does the summand go to zero for $x>1$, as $n $ goes to infinity?
$endgroup$
– AnyAD
Oct 23 '18 at 12:05




$begingroup$
Does the summand go to zero for $x>1$, as $n $ goes to infinity?
$endgroup$
– AnyAD
Oct 23 '18 at 12:05




1




1




$begingroup$
@AnyAD Thanks I understand it now. I have edited the post to show the calculation.
$endgroup$
– StammeringMathematician
Oct 23 '18 at 12:24




$begingroup$
@AnyAD Thanks I understand it now. I have edited the post to show the calculation.
$endgroup$
– StammeringMathematician
Oct 23 '18 at 12:24










1 Answer
1






active

oldest

votes


















2












$begingroup$

Define $$f_n(x)={x^{2n}over n^2+x^{2n}}$$In the interval $[-1,1]$ we have $0le x^{2n}le 1$ therefore $$|f_n(x)|=f_n(x)le {1over n^2+1}$$Now define $M_n={1over n^2+1}$. Obviously the series $sum M_n$ is convergent because$$sum_{n=1}^{infty}M_n=sum_{n=1}^{infty}{1over n^2+1}le sum_{n=1}^{infty}{1over n^2}={pi^2over 6}$$and $$|f_n(x)|le M_n$$which means that the sequence of ${M_n}$ fulfills the conditions of the theorem and using that theorem, we have proved what we wanted.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I think you want $f_n(x)le frac{1}{n^2}.$
    $endgroup$
    – zhw.
    Nov 30 '18 at 16:55










  • $begingroup$
    Also ${1over n^2}$ is another possibility for $M_n$ since ${1over n^2+1}<{1over n^2}$
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 16:57










  • $begingroup$
    But why do you say $$f_n(x)le {1over n^2+1}?$$
    $endgroup$
    – zhw.
    Nov 30 '18 at 17:01










  • $begingroup$
    I'm just looking for some $M_n$ so that the conditions of the theorem hold i.e. $$|f_n(x)|le M_n$$since we have $$|f_n(x)|le {1over n^2}$$ we can arbitrarily choose $M_n={1over n^2+1}$
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 17:05










  • $begingroup$
    But $1/(n^2+1)<1/n^2.$
    $endgroup$
    – zhw.
    Nov 30 '18 at 17:08











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1 Answer
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1 Answer
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active

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active

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active

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2












$begingroup$

Define $$f_n(x)={x^{2n}over n^2+x^{2n}}$$In the interval $[-1,1]$ we have $0le x^{2n}le 1$ therefore $$|f_n(x)|=f_n(x)le {1over n^2+1}$$Now define $M_n={1over n^2+1}$. Obviously the series $sum M_n$ is convergent because$$sum_{n=1}^{infty}M_n=sum_{n=1}^{infty}{1over n^2+1}le sum_{n=1}^{infty}{1over n^2}={pi^2over 6}$$and $$|f_n(x)|le M_n$$which means that the sequence of ${M_n}$ fulfills the conditions of the theorem and using that theorem, we have proved what we wanted.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I think you want $f_n(x)le frac{1}{n^2}.$
    $endgroup$
    – zhw.
    Nov 30 '18 at 16:55










  • $begingroup$
    Also ${1over n^2}$ is another possibility for $M_n$ since ${1over n^2+1}<{1over n^2}$
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 16:57










  • $begingroup$
    But why do you say $$f_n(x)le {1over n^2+1}?$$
    $endgroup$
    – zhw.
    Nov 30 '18 at 17:01










  • $begingroup$
    I'm just looking for some $M_n$ so that the conditions of the theorem hold i.e. $$|f_n(x)|le M_n$$since we have $$|f_n(x)|le {1over n^2}$$ we can arbitrarily choose $M_n={1over n^2+1}$
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 17:05










  • $begingroup$
    But $1/(n^2+1)<1/n^2.$
    $endgroup$
    – zhw.
    Nov 30 '18 at 17:08
















2












$begingroup$

Define $$f_n(x)={x^{2n}over n^2+x^{2n}}$$In the interval $[-1,1]$ we have $0le x^{2n}le 1$ therefore $$|f_n(x)|=f_n(x)le {1over n^2+1}$$Now define $M_n={1over n^2+1}$. Obviously the series $sum M_n$ is convergent because$$sum_{n=1}^{infty}M_n=sum_{n=1}^{infty}{1over n^2+1}le sum_{n=1}^{infty}{1over n^2}={pi^2over 6}$$and $$|f_n(x)|le M_n$$which means that the sequence of ${M_n}$ fulfills the conditions of the theorem and using that theorem, we have proved what we wanted.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I think you want $f_n(x)le frac{1}{n^2}.$
    $endgroup$
    – zhw.
    Nov 30 '18 at 16:55










  • $begingroup$
    Also ${1over n^2}$ is another possibility for $M_n$ since ${1over n^2+1}<{1over n^2}$
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 16:57










  • $begingroup$
    But why do you say $$f_n(x)le {1over n^2+1}?$$
    $endgroup$
    – zhw.
    Nov 30 '18 at 17:01










  • $begingroup$
    I'm just looking for some $M_n$ so that the conditions of the theorem hold i.e. $$|f_n(x)|le M_n$$since we have $$|f_n(x)|le {1over n^2}$$ we can arbitrarily choose $M_n={1over n^2+1}$
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 17:05










  • $begingroup$
    But $1/(n^2+1)<1/n^2.$
    $endgroup$
    – zhw.
    Nov 30 '18 at 17:08














2












2








2





$begingroup$

Define $$f_n(x)={x^{2n}over n^2+x^{2n}}$$In the interval $[-1,1]$ we have $0le x^{2n}le 1$ therefore $$|f_n(x)|=f_n(x)le {1over n^2+1}$$Now define $M_n={1over n^2+1}$. Obviously the series $sum M_n$ is convergent because$$sum_{n=1}^{infty}M_n=sum_{n=1}^{infty}{1over n^2+1}le sum_{n=1}^{infty}{1over n^2}={pi^2over 6}$$and $$|f_n(x)|le M_n$$which means that the sequence of ${M_n}$ fulfills the conditions of the theorem and using that theorem, we have proved what we wanted.






share|cite|improve this answer











$endgroup$



Define $$f_n(x)={x^{2n}over n^2+x^{2n}}$$In the interval $[-1,1]$ we have $0le x^{2n}le 1$ therefore $$|f_n(x)|=f_n(x)le {1over n^2+1}$$Now define $M_n={1over n^2+1}$. Obviously the series $sum M_n$ is convergent because$$sum_{n=1}^{infty}M_n=sum_{n=1}^{infty}{1over n^2+1}le sum_{n=1}^{infty}{1over n^2}={pi^2over 6}$$and $$|f_n(x)|le M_n$$which means that the sequence of ${M_n}$ fulfills the conditions of the theorem and using that theorem, we have proved what we wanted.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 '18 at 17:33

























answered Nov 30 '18 at 16:10









Mostafa AyazMostafa Ayaz

15.6k3939




15.6k3939












  • $begingroup$
    I think you want $f_n(x)le frac{1}{n^2}.$
    $endgroup$
    – zhw.
    Nov 30 '18 at 16:55










  • $begingroup$
    Also ${1over n^2}$ is another possibility for $M_n$ since ${1over n^2+1}<{1over n^2}$
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 16:57










  • $begingroup$
    But why do you say $$f_n(x)le {1over n^2+1}?$$
    $endgroup$
    – zhw.
    Nov 30 '18 at 17:01










  • $begingroup$
    I'm just looking for some $M_n$ so that the conditions of the theorem hold i.e. $$|f_n(x)|le M_n$$since we have $$|f_n(x)|le {1over n^2}$$ we can arbitrarily choose $M_n={1over n^2+1}$
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 17:05










  • $begingroup$
    But $1/(n^2+1)<1/n^2.$
    $endgroup$
    – zhw.
    Nov 30 '18 at 17:08


















  • $begingroup$
    I think you want $f_n(x)le frac{1}{n^2}.$
    $endgroup$
    – zhw.
    Nov 30 '18 at 16:55










  • $begingroup$
    Also ${1over n^2}$ is another possibility for $M_n$ since ${1over n^2+1}<{1over n^2}$
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 16:57










  • $begingroup$
    But why do you say $$f_n(x)le {1over n^2+1}?$$
    $endgroup$
    – zhw.
    Nov 30 '18 at 17:01










  • $begingroup$
    I'm just looking for some $M_n$ so that the conditions of the theorem hold i.e. $$|f_n(x)|le M_n$$since we have $$|f_n(x)|le {1over n^2}$$ we can arbitrarily choose $M_n={1over n^2+1}$
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 17:05










  • $begingroup$
    But $1/(n^2+1)<1/n^2.$
    $endgroup$
    – zhw.
    Nov 30 '18 at 17:08
















$begingroup$
I think you want $f_n(x)le frac{1}{n^2}.$
$endgroup$
– zhw.
Nov 30 '18 at 16:55




$begingroup$
I think you want $f_n(x)le frac{1}{n^2}.$
$endgroup$
– zhw.
Nov 30 '18 at 16:55












$begingroup$
Also ${1over n^2}$ is another possibility for $M_n$ since ${1over n^2+1}<{1over n^2}$
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 16:57




$begingroup$
Also ${1over n^2}$ is another possibility for $M_n$ since ${1over n^2+1}<{1over n^2}$
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 16:57












$begingroup$
But why do you say $$f_n(x)le {1over n^2+1}?$$
$endgroup$
– zhw.
Nov 30 '18 at 17:01




$begingroup$
But why do you say $$f_n(x)le {1over n^2+1}?$$
$endgroup$
– zhw.
Nov 30 '18 at 17:01












$begingroup$
I'm just looking for some $M_n$ so that the conditions of the theorem hold i.e. $$|f_n(x)|le M_n$$since we have $$|f_n(x)|le {1over n^2}$$ we can arbitrarily choose $M_n={1over n^2+1}$
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 17:05




$begingroup$
I'm just looking for some $M_n$ so that the conditions of the theorem hold i.e. $$|f_n(x)|le M_n$$since we have $$|f_n(x)|le {1over n^2}$$ we can arbitrarily choose $M_n={1over n^2+1}$
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 17:05












$begingroup$
But $1/(n^2+1)<1/n^2.$
$endgroup$
– zhw.
Nov 30 '18 at 17:08




$begingroup$
But $1/(n^2+1)<1/n^2.$
$endgroup$
– zhw.
Nov 30 '18 at 17:08


















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