Evaluate $sum_r=1^infty sqrt frac rr^4+r^2+1$
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One of friend gave me a question today to solve which is as follows
$$sum_r=1^infty sqrt frac rr^4+r^2+1$$
In spite of much efforts I couldn't solve it and so I asked him to check whether the question was correct or was it this one
$$sum_r=1^infty frac rr^4+r^2+1$$
I thought the question would be this one because the terms inside the root can be telescoped in absence of root. And indeed I was right. The question was as I expected the latter one.
But even after that I thought about whether the first question containing the square root could also be solved or not. For just checking out the convergence of the sequence I tried the ratio test but it wasn't quite helpful. Then I tried using the integral test and indeed $$lim_tto infty int_1^t sqrtfrac xx^4+x^2+1 dx$$
this integral converges to $1.80984$ according to Wolfy. Upon lot of efforts too I am not an able to solve the first summation (with the square root terms) . Can someone please lend me some help over this problem.
calculus integration sequences-and-series limits convergence
asked Jun 5 at 15:28
Manthanein
6,1541437
 |Â
show 2 more comments
up vote
7
down vote
favorite
One of friend gave me a question today to solve which is as follows
$$sum_r=1^infty sqrt frac rr^4+r^2+1$$
In spite of much efforts I couldn't solve it and so I asked him to check whether the question was correct or was it this one
$$sum_r=1^infty frac rr^4+r^2+1$$
I thought the question would be this one because the terms inside the root can be telescoped in absence of root. And indeed I was right. The question was as I expected the latter one.
But even after that I thought about whether the first question containing the square root could also be solved or not. For just checking out the convergence of the sequence I tried the ratio test but it wasn't quite helpful. Then I tried using the integral test and indeed $$lim_tto infty int_1^t sqrtfrac xx^4+x^2+1 dx$$
this integral converges to $1.80984$ according to Wolfy. Upon lot of efforts too I am not an able to solve the first summation (with the square root terms) . Can someone please lend me some help over this problem.
calculus integration sequences-and-series limits convergence
asked Jun 5 at 15:28
Manthanein
6,1541437
You can try to rationalize the denominator to get rid of the square root. Then use partial fractions...
– Frank W.
Jun 5 at 15:42
@Frank W. What should I take the rationalizing factor as? Cause I don't see anything popping up
– Manthanein
Jun 5 at 15:44
Yeah, I don't think rationalizing the denominator is gonna work here. @FrankW. You probably aren't going to be able to get a telescoping series because partial sums are a bit mad, and telescoping sequences have nice partial sums.
– Thomas Andrews
Jun 5 at 15:55
1
@ThomasAndrews Yes, I tried that route. Just went uglier and uglier. If you enter it into Wolfram Alpha, you get a decimal approximation. So most likely, there isn't quite a closed form for this sum.
– Frank W.
Jun 5 at 16:13
@Manthanein I was thinking that you could multiply the denominator and numerator by $(r^4+r^2+1)^1/2$ so the denominator is free of any square roots. Didn't work though, the result was still rather ugly.
– Frank W.
Jun 5 at 16:13
 |Â
show 2 more comments
up vote
7
down vote
favorite
up vote
7
down vote
favorite
One of friend gave me a question today to solve which is as follows
$$sum_r=1^infty sqrt frac rr^4+r^2+1$$
In spite of much efforts I couldn't solve it and so I asked him to check whether the question was correct or was it this one
$$sum_r=1^infty frac rr^4+r^2+1$$
I thought the question would be this one because the terms inside the root can be telescoped in absence of root. And indeed I was right. The question was as I expected the latter one.
But even after that I thought about whether the first question containing the square root could also be solved or not. For just checking out the convergence of the sequence I tried the ratio test but it wasn't quite helpful. Then I tried using the integral test and indeed $$lim_tto infty int_1^t sqrtfrac xx^4+x^2+1 dx$$
this integral converges to $1.80984$ according to Wolfy. Upon lot of efforts too I am not an able to solve the first summation (with the square root terms) . Can someone please lend me some help over this problem.
calculus integration sequences-and-series limits convergence
asked Jun 5 at 15:28
Manthanein
6,1541437
One of friend gave me a question today to solve which is as follows
$$sum_r=1^infty sqrt frac rr^4+r^2+1$$
In spite of much efforts I couldn't solve it and so I asked him to check whether the question was correct or was it this one
$$sum_r=1^infty frac rr^4+r^2+1$$
I thought the question would be this one because the terms inside the root can be telescoped in absence of root. And indeed I was right. The question was as I expected the latter one.
But even after that I thought about whether the first question containing the square root could also be solved or not. For just checking out the convergence of the sequence I tried the ratio test but it wasn't quite helpful. Then I tried using the integral test and indeed $$lim_tto infty int_1^t sqrtfrac xx^4+x^2+1 dx$$
this integral converges to $1.80984$ according to Wolfy. Upon lot of efforts too I am not an able to solve the first summation (with the square root terms) . Can someone please lend me some help over this problem.
calculus integration sequences-and-series limits convergence
asked Jun 5 at 15:28
Manthanein
6,1541437
edited Aug 9 at 19:16
asked Jun 5 at 15:28
Manthanein
6,1541437
asked Jun 5 at 15:28
Manthanein
6,1541437
asked Jun 5 at 15:28
Manthanein
6,1541437
6,1541437
You can try to rationalize the denominator to get rid of the square root. Then use partial fractions...
– Frank W.
Jun 5 at 15:42
@Frank W. What should I take the rationalizing factor as? Cause I don't see anything popping up
– Manthanein
Jun 5 at 15:44
Yeah, I don't think rationalizing the denominator is gonna work here. @FrankW. You probably aren't going to be able to get a telescoping series because partial sums are a bit mad, and telescoping sequences have nice partial sums.
– Thomas Andrews
Jun 5 at 15:55
1
@ThomasAndrews Yes, I tried that route. Just went uglier and uglier. If you enter it into Wolfram Alpha, you get a decimal approximation. So most likely, there isn't quite a closed form for this sum.
– Frank W.
Jun 5 at 16:13
@Manthanein I was thinking that you could multiply the denominator and numerator by $(r^4+r^2+1)^1/2$ so the denominator is free of any square roots. Didn't work though, the result was still rather ugly.
– Frank W.
Jun 5 at 16:13
 |Â
show 2 more comments
You can try to rationalize the denominator to get rid of the square root. Then use partial fractions...
– Frank W.
Jun 5 at 15:42
@Frank W. What should I take the rationalizing factor as? Cause I don't see anything popping up
– Manthanein
Jun 5 at 15:44
Yeah, I don't think rationalizing the denominator is gonna work here. @FrankW. You probably aren't going to be able to get a telescoping series because partial sums are a bit mad, and telescoping sequences have nice partial sums.
– Thomas Andrews
Jun 5 at 15:55
1
@ThomasAndrews Yes, I tried that route. Just went uglier and uglier. If you enter it into Wolfram Alpha, you get a decimal approximation. So most likely, there isn't quite a closed form for this sum.
– Frank W.
Jun 5 at 16:13
@Manthanein I was thinking that you could multiply the denominator and numerator by $(r^4+r^2+1)^1/2$ so the denominator is free of any square roots. Didn't work though, the result was still rather ugly.
– Frank W.
Jun 5 at 16:13
You can try to rationalize the denominator to get rid of the square root. Then use partial fractions...
– Frank W.
Jun 5 at 15:42
You can try to rationalize the denominator to get rid of the square root. Then use partial fractions...
– Frank W.
Jun 5 at 15:42
@Frank W. What should I take the rationalizing factor as? Cause I don't see anything popping up
– Manthanein
Jun 5 at 15:44
@Frank W. What should I take the rationalizing factor as? Cause I don't see anything popping up
– Manthanein
Jun 5 at 15:44
Yeah, I don't think rationalizing the denominator is gonna work here. @FrankW. You probably aren't going to be able to get a telescoping series because partial sums are a bit mad, and telescoping sequences have nice partial sums.
– Thomas Andrews
Jun 5 at 15:55
Yeah, I don't think rationalizing the denominator is gonna work here. @FrankW. You probably aren't going to be able to get a telescoping series because partial sums are a bit mad, and telescoping sequences have nice partial sums.
– Thomas Andrews
Jun 5 at 15:55
1
1
@ThomasAndrews Yes, I tried that route. Just went uglier and uglier. If you enter it into Wolfram Alpha, you get a decimal approximation. So most likely, there isn't quite a closed form for this sum.
– Frank W.
Jun 5 at 16:13
@ThomasAndrews Yes, I tried that route. Just went uglier and uglier. If you enter it into Wolfram Alpha, you get a decimal approximation. So most likely, there isn't quite a closed form for this sum.
– Frank W.
Jun 5 at 16:13
@Manthanein I was thinking that you could multiply the denominator and numerator by $(r^4+r^2+1)^1/2$ so the denominator is free of any square roots. Didn't work though, the result was still rather ugly.
– Frank W.
Jun 5 at 16:13
@Manthanein I was thinking that you could multiply the denominator and numerator by $(r^4+r^2+1)^1/2$ so the denominator is free of any square roots. Didn't work though, the result was still rather ugly.
– Frank W.
Jun 5 at 16:13
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
We have $$sqrtfracrr^4+r^2+1approx frac2sqrtr-1/2-frac2sqrtr+1/2,$$ so by creative telescoping the value of the series is not too far from $frac1+2sqrt2sqrt3$, and this approach can be improved in order to obtain more accurate approximations. However I won't bet on a simple closed form for the given series. Numerical methods produce $approx 2.12534074896$, which not by chance is pretty close to
$$ int_1/2^+inftysqrtfracxx^4+x^2+1,dxapprox 2.12, $$
an incomplete elliptic integral.
answered Aug 9 at 20:54
Jack D'Aurizio♦
271k31266631
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
We have $$sqrtfracrr^4+r^2+1approx frac2sqrtr-1/2-frac2sqrtr+1/2,$$ so by creative telescoping the value of the series is not too far from $frac1+2sqrt2sqrt3$, and this approach can be improved in order to obtain more accurate approximations. However I won't bet on a simple closed form for the given series. Numerical methods produce $approx 2.12534074896$, which not by chance is pretty close to
$$ int_1/2^+inftysqrtfracxx^4+x^2+1,dxapprox 2.12, $$
an incomplete elliptic integral.
answered Aug 9 at 20:54
Jack D'Aurizio♦
271k31266631
add a comment |Â
up vote
4
down vote
accepted
We have $$sqrtfracrr^4+r^2+1approx frac2sqrtr-1/2-frac2sqrtr+1/2,$$ so by creative telescoping the value of the series is not too far from $frac1+2sqrt2sqrt3$, and this approach can be improved in order to obtain more accurate approximations. However I won't bet on a simple closed form for the given series. Numerical methods produce $approx 2.12534074896$, which not by chance is pretty close to
$$ int_1/2^+inftysqrtfracxx^4+x^2+1,dxapprox 2.12, $$
an incomplete elliptic integral.
answered Aug 9 at 20:54
Jack D'Aurizio♦
271k31266631
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
We have $$sqrtfracrr^4+r^2+1approx frac2sqrtr-1/2-frac2sqrtr+1/2,$$ so by creative telescoping the value of the series is not too far from $frac1+2sqrt2sqrt3$, and this approach can be improved in order to obtain more accurate approximations. However I won't bet on a simple closed form for the given series. Numerical methods produce $approx 2.12534074896$, which not by chance is pretty close to
$$ int_1/2^+inftysqrtfracxx^4+x^2+1,dxapprox 2.12, $$
an incomplete elliptic integral.
answered Aug 9 at 20:54
Jack D'Aurizio♦
271k31266631
We have $$sqrtfracrr^4+r^2+1approx frac2sqrtr-1/2-frac2sqrtr+1/2,$$ so by creative telescoping the value of the series is not too far from $frac1+2sqrt2sqrt3$, and this approach can be improved in order to obtain more accurate approximations. However I won't bet on a simple closed form for the given series. Numerical methods produce $approx 2.12534074896$, which not by chance is pretty close to
$$ int_1/2^+inftysqrtfracxx^4+x^2+1,dxapprox 2.12, $$
an incomplete elliptic integral.
answered Aug 9 at 20:54
Jack D'Aurizio♦
271k31266631
answered Aug 9 at 20:54
Jack D'Aurizio♦
271k31266631
answered Aug 9 at 20:54
Jack D'Aurizio♦
271k31266631
answered Aug 9 at 20:54
Jack D'Aurizio♦
271k31266631
271k31266631
add a comment |Â
add a comment |Â
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You can try to rationalize the denominator to get rid of the square root. Then use partial fractions...
– Frank W.
Jun 5 at 15:42
@Frank W. What should I take the rationalizing factor as? Cause I don't see anything popping up
– Manthanein
Jun 5 at 15:44
Yeah, I don't think rationalizing the denominator is gonna work here. @FrankW. You probably aren't going to be able to get a telescoping series because partial sums are a bit mad, and telescoping sequences have nice partial sums.
– Thomas Andrews
Jun 5 at 15:55
1
@ThomasAndrews Yes, I tried that route. Just went uglier and uglier. If you enter it into Wolfram Alpha, you get a decimal approximation. So most likely, there isn't quite a closed form for this sum.
– Frank W.
Jun 5 at 16:13
@Manthanein I was thinking that you could multiply the denominator and numerator by $(r^4+r^2+1)^1/2$ so the denominator is free of any square roots. Didn't work though, the result was still rather ugly.
– Frank W.
Jun 5 at 16:13