Study the convergence of the function series $sum_{n=1}^{infty}arctan(frac{2x}{x^2+n^4})$











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0
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I need to study the convergence of the series $sum_{n=1}^{infty}arctan(frac{2x}{x^2+n^4})$.



I thought about 2 ways it can be done.



One way is to try and calculate the partial sum of the series so we can see what is its limit.



Another way is to use the fact that if we have $sum_{n=1}^{infty}f_n(x)$ and we find a sequence $a_n$ so that



$|f_n(x)| leq a_n$



and



$sum_{n=1}^{infty}a_n$ is convergent



then $sum_{n=1}^{infty}f_n(x)$ is convergent.



Now, we know that $arctan : mathbb{R} -> (-frac{pi}{2}, frac{pi}{2})$ but if we take $a_n = frac{pi}{2}$ then $sum_{n=1}^{infty}a_n$ is divergent, but maybe there is a sequence that is bigger than $|arctan(frac{2x}{x^2+n^4})|$ but it is convergent.



Can you help me out to see how to study the convergence of this function series?






sequences-and-series







share|cite|improve this question
























  • Note that $arctan$ is increasing and $frac{2x}{x^2+n^4}$ has a maximum for $x=n^2$. This allows you determine if the convergence is uniform or not.
    – Winther
    Nov 21 at 22:05












  • @Winther How do I show that, let's say $x = n^2$ is a maximum?I think I need to study the monotony of the function to show that, right?
    – Ghost
    Nov 21 at 22:11










  • For example take derivative $=0$ to find extremal points. Also note that if you show $|f_n(x)| leq a_n$ for some $a_n$ where $sum a_n$ converges then you don't only show that $sum f_n(x)$ is convergent, but that the convergence is uniform (this is Weierstrass M-test)
    – Winther
    Nov 21 at 22:12












  • @Winther Yes, that's what I thought about. Thanks for the tip.
    – Ghost
    Nov 21 at 22:13















up vote
0
down vote

favorite












I need to study the convergence of the series $sum_{n=1}^{infty}arctan(frac{2x}{x^2+n^4})$.



I thought about 2 ways it can be done.



One way is to try and calculate the partial sum of the series so we can see what is its limit.



Another way is to use the fact that if we have $sum_{n=1}^{infty}f_n(x)$ and we find a sequence $a_n$ so that



$|f_n(x)| leq a_n$



and



$sum_{n=1}^{infty}a_n$ is convergent



then $sum_{n=1}^{infty}f_n(x)$ is convergent.



Now, we know that $arctan : mathbb{R} -> (-frac{pi}{2}, frac{pi}{2})$ but if we take $a_n = frac{pi}{2}$ then $sum_{n=1}^{infty}a_n$ is divergent, but maybe there is a sequence that is bigger than $|arctan(frac{2x}{x^2+n^4})|$ but it is convergent.



Can you help me out to see how to study the convergence of this function series?






sequences-and-series







share|cite|improve this question
























  • Note that $arctan$ is increasing and $frac{2x}{x^2+n^4}$ has a maximum for $x=n^2$. This allows you determine if the convergence is uniform or not.
    – Winther
    Nov 21 at 22:05












  • @Winther How do I show that, let's say $x = n^2$ is a maximum?I think I need to study the monotony of the function to show that, right?
    – Ghost
    Nov 21 at 22:11










  • For example take derivative $=0$ to find extremal points. Also note that if you show $|f_n(x)| leq a_n$ for some $a_n$ where $sum a_n$ converges then you don't only show that $sum f_n(x)$ is convergent, but that the convergence is uniform (this is Weierstrass M-test)
    – Winther
    Nov 21 at 22:12












  • @Winther Yes, that's what I thought about. Thanks for the tip.
    – Ghost
    Nov 21 at 22:13













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I need to study the convergence of the series $sum_{n=1}^{infty}arctan(frac{2x}{x^2+n^4})$.



I thought about 2 ways it can be done.



One way is to try and calculate the partial sum of the series so we can see what is its limit.



Another way is to use the fact that if we have $sum_{n=1}^{infty}f_n(x)$ and we find a sequence $a_n$ so that



$|f_n(x)| leq a_n$



and



$sum_{n=1}^{infty}a_n$ is convergent



then $sum_{n=1}^{infty}f_n(x)$ is convergent.



Now, we know that $arctan : mathbb{R} -> (-frac{pi}{2}, frac{pi}{2})$ but if we take $a_n = frac{pi}{2}$ then $sum_{n=1}^{infty}a_n$ is divergent, but maybe there is a sequence that is bigger than $|arctan(frac{2x}{x^2+n^4})|$ but it is convergent.



Can you help me out to see how to study the convergence of this function series?






sequences-and-series







share|cite|improve this question















I need to study the convergence of the series $sum_{n=1}^{infty}arctan(frac{2x}{x^2+n^4})$.



I thought about 2 ways it can be done.



One way is to try and calculate the partial sum of the series so we can see what is its limit.



Another way is to use the fact that if we have $sum_{n=1}^{infty}f_n(x)$ and we find a sequence $a_n$ so that



$|f_n(x)| leq a_n$



and



$sum_{n=1}^{infty}a_n$ is convergent



then $sum_{n=1}^{infty}f_n(x)$ is convergent.



Now, we know that $arctan : mathbb{R} -> (-frac{pi}{2}, frac{pi}{2})$ but if we take $a_n = frac{pi}{2}$ then $sum_{n=1}^{infty}a_n$ is divergent, but maybe there is a sequence that is bigger than $|arctan(frac{2x}{x^2+n^4})|$ but it is convergent.



Can you help me out to see how to study the convergence of this function series?






sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 at 22:00









gimusi

91.9k84495




91.9k84495










asked Nov 21 at 21:54









Ghost

553210




553210












  • Note that $arctan$ is increasing and $frac{2x}{x^2+n^4}$ has a maximum for $x=n^2$. This allows you determine if the convergence is uniform or not.
    – Winther
    Nov 21 at 22:05












  • @Winther How do I show that, let's say $x = n^2$ is a maximum?I think I need to study the monotony of the function to show that, right?
    – Ghost
    Nov 21 at 22:11










  • For example take derivative $=0$ to find extremal points. Also note that if you show $|f_n(x)| leq a_n$ for some $a_n$ where $sum a_n$ converges then you don't only show that $sum f_n(x)$ is convergent, but that the convergence is uniform (this is Weierstrass M-test)
    – Winther
    Nov 21 at 22:12












  • @Winther Yes, that's what I thought about. Thanks for the tip.
    – Ghost
    Nov 21 at 22:13


















  • Note that $arctan$ is increasing and $frac{2x}{x^2+n^4}$ has a maximum for $x=n^2$. This allows you determine if the convergence is uniform or not.
    – Winther
    Nov 21 at 22:05












  • @Winther How do I show that, let's say $x = n^2$ is a maximum?I think I need to study the monotony of the function to show that, right?
    – Ghost
    Nov 21 at 22:11










  • For example take derivative $=0$ to find extremal points. Also note that if you show $|f_n(x)| leq a_n$ for some $a_n$ where $sum a_n$ converges then you don't only show that $sum f_n(x)$ is convergent, but that the convergence is uniform (this is Weierstrass M-test)
    – Winther
    Nov 21 at 22:12












  • @Winther Yes, that's what I thought about. Thanks for the tip.
    – Ghost
    Nov 21 at 22:13
















Note that $arctan$ is increasing and $frac{2x}{x^2+n^4}$ has a maximum for $x=n^2$. This allows you determine if the convergence is uniform or not.
– Winther
Nov 21 at 22:05






Note that $arctan$ is increasing and $frac{2x}{x^2+n^4}$ has a maximum for $x=n^2$. This allows you determine if the convergence is uniform or not.
– Winther
Nov 21 at 22:05














@Winther How do I show that, let's say $x = n^2$ is a maximum?I think I need to study the monotony of the function to show that, right?
– Ghost
Nov 21 at 22:11




@Winther How do I show that, let's say $x = n^2$ is a maximum?I think I need to study the monotony of the function to show that, right?
– Ghost
Nov 21 at 22:11












For example take derivative $=0$ to find extremal points. Also note that if you show $|f_n(x)| leq a_n$ for some $a_n$ where $sum a_n$ converges then you don't only show that $sum f_n(x)$ is convergent, but that the convergence is uniform (this is Weierstrass M-test)
– Winther
Nov 21 at 22:12






For example take derivative $=0$ to find extremal points. Also note that if you show $|f_n(x)| leq a_n$ for some $a_n$ where $sum a_n$ converges then you don't only show that $sum f_n(x)$ is convergent, but that the convergence is uniform (this is Weierstrass M-test)
– Winther
Nov 21 at 22:12














@Winther Yes, that's what I thought about. Thanks for the tip.
– Ghost
Nov 21 at 22:13




@Winther Yes, that's what I thought about. Thanks for the tip.
– Ghost
Nov 21 at 22:13










2 Answers
2






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2
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Hint: For $x ge 0$, Using AM-GM inequality we have: $x^2+n^4 ge 2xn^2implies dfrac{2x}{x^2+n^4}le dfrac{1}{n^2}implies tan^{-1}left(frac{2x}{x^2+n^4}right)le tan^{-1}left(frac{1}{n^2}right)ledfrac{1}{n^2}$ ( note: $tan^{-1}(theta) le theta, theta ge 0$ ,and $tan^{-1}(theta)$ is a increasing function of $theta$. For $x < 0$, use $tan^{-1}(-x) = -tan^{-1}(x)$.And the comparison test shows the series convergent.






share|cite|improve this answer





















  • Do I need to take 2 separate cases, for $x<0$ and $x>0$? I think that if I use absolute value I do not need to.
    – Ghost
    Nov 21 at 22:16




















up vote
0
down vote













We have that for any $xneq 0$



$$arctanleft(frac{2x}{x^2+n^4}right) sim frac{2x}{n^4}$$



therefore the series converges by limit comparison test with $sum frac1{n^2}$.






share|cite|improve this answer





















  • I thought about the fact that $arctan(frac{2x}{x^2+n^4}) leq arctan(frac{2x}{n^4})$, but then again, why $arctan(frac{2x}{x^2+n^4}) sim frac{2x}{n^4}$? Are you using the fact that if $lim_{n->infty}frac{a_n}{b_n}$ exists then $a_n$ and $b_n$ have the same nature?
    – Ghost
    Nov 21 at 22:03












  • @Ghost We are using that as $y to 0 quad frac{arctan y}{y}to 1$ and then imit comparison test.
    – gimusi
    Nov 21 at 22:06












  • @Ghost That is for $n$ large $arctanleft(frac{2x}{x^2+n^4}right) $ behaves as $frac{2x}{x^2+n^4}$ and the latter converges by limit comparison test with $sum frac1{n^2}$.
    – gimusi
    Nov 21 at 22:07













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2 Answers
2






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2 Answers
2






active

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active

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active

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up vote
2
down vote













Hint: For $x ge 0$, Using AM-GM inequality we have: $x^2+n^4 ge 2xn^2implies dfrac{2x}{x^2+n^4}le dfrac{1}{n^2}implies tan^{-1}left(frac{2x}{x^2+n^4}right)le tan^{-1}left(frac{1}{n^2}right)ledfrac{1}{n^2}$ ( note: $tan^{-1}(theta) le theta, theta ge 0$ ,and $tan^{-1}(theta)$ is a increasing function of $theta$. For $x < 0$, use $tan^{-1}(-x) = -tan^{-1}(x)$.And the comparison test shows the series convergent.






share|cite|improve this answer





















  • Do I need to take 2 separate cases, for $x<0$ and $x>0$? I think that if I use absolute value I do not need to.
    – Ghost
    Nov 21 at 22:16

















up vote
2
down vote













Hint: For $x ge 0$, Using AM-GM inequality we have: $x^2+n^4 ge 2xn^2implies dfrac{2x}{x^2+n^4}le dfrac{1}{n^2}implies tan^{-1}left(frac{2x}{x^2+n^4}right)le tan^{-1}left(frac{1}{n^2}right)ledfrac{1}{n^2}$ ( note: $tan^{-1}(theta) le theta, theta ge 0$ ,and $tan^{-1}(theta)$ is a increasing function of $theta$. For $x < 0$, use $tan^{-1}(-x) = -tan^{-1}(x)$.And the comparison test shows the series convergent.






share|cite|improve this answer





















  • Do I need to take 2 separate cases, for $x<0$ and $x>0$? I think that if I use absolute value I do not need to.
    – Ghost
    Nov 21 at 22:16















up vote
2
down vote










up vote
2
down vote









Hint: For $x ge 0$, Using AM-GM inequality we have: $x^2+n^4 ge 2xn^2implies dfrac{2x}{x^2+n^4}le dfrac{1}{n^2}implies tan^{-1}left(frac{2x}{x^2+n^4}right)le tan^{-1}left(frac{1}{n^2}right)ledfrac{1}{n^2}$ ( note: $tan^{-1}(theta) le theta, theta ge 0$ ,and $tan^{-1}(theta)$ is a increasing function of $theta$. For $x < 0$, use $tan^{-1}(-x) = -tan^{-1}(x)$.And the comparison test shows the series convergent.






share|cite|improve this answer












Hint: For $x ge 0$, Using AM-GM inequality we have: $x^2+n^4 ge 2xn^2implies dfrac{2x}{x^2+n^4}le dfrac{1}{n^2}implies tan^{-1}left(frac{2x}{x^2+n^4}right)le tan^{-1}left(frac{1}{n^2}right)ledfrac{1}{n^2}$ ( note: $tan^{-1}(theta) le theta, theta ge 0$ ,and $tan^{-1}(theta)$ is a increasing function of $theta$. For $x < 0$, use $tan^{-1}(-x) = -tan^{-1}(x)$.And the comparison test shows the series convergent.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 21 at 22:10









DeepSea

70.8k54487




70.8k54487












  • Do I need to take 2 separate cases, for $x<0$ and $x>0$? I think that if I use absolute value I do not need to.
    – Ghost
    Nov 21 at 22:16




















  • Do I need to take 2 separate cases, for $x<0$ and $x>0$? I think that if I use absolute value I do not need to.
    – Ghost
    Nov 21 at 22:16


















Do I need to take 2 separate cases, for $x<0$ and $x>0$? I think that if I use absolute value I do not need to.
– Ghost
Nov 21 at 22:16






Do I need to take 2 separate cases, for $x<0$ and $x>0$? I think that if I use absolute value I do not need to.
– Ghost
Nov 21 at 22:16












up vote
0
down vote













We have that for any $xneq 0$



$$arctanleft(frac{2x}{x^2+n^4}right) sim frac{2x}{n^4}$$



therefore the series converges by limit comparison test with $sum frac1{n^2}$.






share|cite|improve this answer





















  • I thought about the fact that $arctan(frac{2x}{x^2+n^4}) leq arctan(frac{2x}{n^4})$, but then again, why $arctan(frac{2x}{x^2+n^4}) sim frac{2x}{n^4}$? Are you using the fact that if $lim_{n->infty}frac{a_n}{b_n}$ exists then $a_n$ and $b_n$ have the same nature?
    – Ghost
    Nov 21 at 22:03












  • @Ghost We are using that as $y to 0 quad frac{arctan y}{y}to 1$ and then imit comparison test.
    – gimusi
    Nov 21 at 22:06












  • @Ghost That is for $n$ large $arctanleft(frac{2x}{x^2+n^4}right) $ behaves as $frac{2x}{x^2+n^4}$ and the latter converges by limit comparison test with $sum frac1{n^2}$.
    – gimusi
    Nov 21 at 22:07

















up vote
0
down vote













We have that for any $xneq 0$



$$arctanleft(frac{2x}{x^2+n^4}right) sim frac{2x}{n^4}$$



therefore the series converges by limit comparison test with $sum frac1{n^2}$.






share|cite|improve this answer





















  • I thought about the fact that $arctan(frac{2x}{x^2+n^4}) leq arctan(frac{2x}{n^4})$, but then again, why $arctan(frac{2x}{x^2+n^4}) sim frac{2x}{n^4}$? Are you using the fact that if $lim_{n->infty}frac{a_n}{b_n}$ exists then $a_n$ and $b_n$ have the same nature?
    – Ghost
    Nov 21 at 22:03












  • @Ghost We are using that as $y to 0 quad frac{arctan y}{y}to 1$ and then imit comparison test.
    – gimusi
    Nov 21 at 22:06












  • @Ghost That is for $n$ large $arctanleft(frac{2x}{x^2+n^4}right) $ behaves as $frac{2x}{x^2+n^4}$ and the latter converges by limit comparison test with $sum frac1{n^2}$.
    – gimusi
    Nov 21 at 22:07















up vote
0
down vote










up vote
0
down vote









We have that for any $xneq 0$



$$arctanleft(frac{2x}{x^2+n^4}right) sim frac{2x}{n^4}$$



therefore the series converges by limit comparison test with $sum frac1{n^2}$.






share|cite|improve this answer












We have that for any $xneq 0$



$$arctanleft(frac{2x}{x^2+n^4}right) sim frac{2x}{n^4}$$



therefore the series converges by limit comparison test with $sum frac1{n^2}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 21 at 21:58









gimusi

91.9k84495




91.9k84495












  • I thought about the fact that $arctan(frac{2x}{x^2+n^4}) leq arctan(frac{2x}{n^4})$, but then again, why $arctan(frac{2x}{x^2+n^4}) sim frac{2x}{n^4}$? Are you using the fact that if $lim_{n->infty}frac{a_n}{b_n}$ exists then $a_n$ and $b_n$ have the same nature?
    – Ghost
    Nov 21 at 22:03












  • @Ghost We are using that as $y to 0 quad frac{arctan y}{y}to 1$ and then imit comparison test.
    – gimusi
    Nov 21 at 22:06












  • @Ghost That is for $n$ large $arctanleft(frac{2x}{x^2+n^4}right) $ behaves as $frac{2x}{x^2+n^4}$ and the latter converges by limit comparison test with $sum frac1{n^2}$.
    – gimusi
    Nov 21 at 22:07




















  • I thought about the fact that $arctan(frac{2x}{x^2+n^4}) leq arctan(frac{2x}{n^4})$, but then again, why $arctan(frac{2x}{x^2+n^4}) sim frac{2x}{n^4}$? Are you using the fact that if $lim_{n->infty}frac{a_n}{b_n}$ exists then $a_n$ and $b_n$ have the same nature?
    – Ghost
    Nov 21 at 22:03












  • @Ghost We are using that as $y to 0 quad frac{arctan y}{y}to 1$ and then imit comparison test.
    – gimusi
    Nov 21 at 22:06












  • @Ghost That is for $n$ large $arctanleft(frac{2x}{x^2+n^4}right) $ behaves as $frac{2x}{x^2+n^4}$ and the latter converges by limit comparison test with $sum frac1{n^2}$.
    – gimusi
    Nov 21 at 22:07


















I thought about the fact that $arctan(frac{2x}{x^2+n^4}) leq arctan(frac{2x}{n^4})$, but then again, why $arctan(frac{2x}{x^2+n^4}) sim frac{2x}{n^4}$? Are you using the fact that if $lim_{n->infty}frac{a_n}{b_n}$ exists then $a_n$ and $b_n$ have the same nature?
– Ghost
Nov 21 at 22:03






I thought about the fact that $arctan(frac{2x}{x^2+n^4}) leq arctan(frac{2x}{n^4})$, but then again, why $arctan(frac{2x}{x^2+n^4}) sim frac{2x}{n^4}$? Are you using the fact that if $lim_{n->infty}frac{a_n}{b_n}$ exists then $a_n$ and $b_n$ have the same nature?
– Ghost
Nov 21 at 22:03














@Ghost We are using that as $y to 0 quad frac{arctan y}{y}to 1$ and then imit comparison test.
– gimusi
Nov 21 at 22:06






@Ghost We are using that as $y to 0 quad frac{arctan y}{y}to 1$ and then imit comparison test.
– gimusi
Nov 21 at 22:06














@Ghost That is for $n$ large $arctanleft(frac{2x}{x^2+n^4}right) $ behaves as $frac{2x}{x^2+n^4}$ and the latter converges by limit comparison test with $sum frac1{n^2}$.
– gimusi
Nov 21 at 22:07






@Ghost That is for $n$ large $arctanleft(frac{2x}{x^2+n^4}right) $ behaves as $frac{2x}{x^2+n^4}$ and the latter converges by limit comparison test with $sum frac1{n^2}$.
– gimusi
Nov 21 at 22:07




















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Ellipse (mathématiques)

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Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
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Quarter-circle Tiles

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up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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