Study the convergence of the function series $sum_{n=1}^{infty}arctan(frac{2x}{x^2+n^4})$
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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
edited Nov 21 at 22:00
gimusi
91.9k84495
asked Nov 21 at 21:54
Ghost
553210
add a comment |
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
edited Nov 21 at 22:00
gimusi
91.9k84495
asked Nov 21 at 21:54
Ghost
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
add a comment |
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
edited Nov 21 at 22:00
gimusi
91.9k84495
asked Nov 21 at 21:54
Ghost
553210
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
edited Nov 21 at 22:00
gimusi
91.9k84495
asked Nov 21 at 21:54
Ghost
553210
edited Nov 21 at 22:00
gimusi
91.9k84495
asked Nov 21 at 21:54
Ghost
553210
edited Nov 21 at 22:00
gimusi
91.9k84495
edited Nov 21 at 22:00
gimusi
91.9k84495
edited Nov 21 at 22:00
gimusi
91.9k84495
91.9k84495
asked Nov 21 at 21:54
Ghost
553210
asked Nov 21 at 21:54
Ghost
553210
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
add a comment |
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
add a comment |
2 Answers
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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.
answered Nov 21 at 22:10
DeepSea
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
add a comment |
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 $sum frac1{n^2}$.
answered Nov 21 at 21:58
gimusi
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
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Nov 21 at 22:10
DeepSea
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
add a comment |
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.
answered Nov 21 at 22:10
DeepSea
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
add a comment |
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.
answered Nov 21 at 22:10
DeepSea
70.8k54487
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.
answered Nov 21 at 22:10
DeepSea
70.8k54487
answered Nov 21 at 22:10
DeepSea
70.8k54487
answered Nov 21 at 22:10
DeepSea
70.8k54487
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
add a comment |
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
add a comment |
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 $sum frac1{n^2}$.
answered Nov 21 at 21:58
gimusi
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
add a comment |
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 $sum frac1{n^2}$.
answered Nov 21 at 21:58
gimusi
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
add a comment |
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 $sum frac1{n^2}$.
answered Nov 21 at 21:58
gimusi
91.9k84495
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 $sum frac1{n^2}$.
answered Nov 21 at 21:58
gimusi
91.9k84495
answered Nov 21 at 21:58
gimusi
91.9k84495
answered Nov 21 at 21:58
gimusi
91.9k84495
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
add a comment |
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
add a comment |
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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