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Prove geometrically that $sumlimits_n=1^infty,dfrac1n(n+1)=1$.
Clash Royale CLAN TAG#URR8PPP
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$mathrmS_1$ and $mathrmS_2$ are two circles of unit radius touching at point $P$; $T$ is a common tangent to $mathrmS_1$ and $mathrmS_2$ touching them at $X$ and $Y$; $mathrmC_1$ is the circle touching $mathrmS_1$, $mathrmS_2$, and $T$. $mathrmC_n$ is the circle touching $mathrmS_1$ , $mathrmS_2$, and $mathrmC_n-1$ for $n > 1.$ By computing diameters of $mathrmC_n$ prove that $$frac11cdot 2+frac12cdot 3+frac13cdot 4+ldots =1,.$$
sequences-and-series geometry convergence circle tangent-line
edited Aug 3 at 10:16
Batominovski
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asked Aug 2 at 14:50
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808
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$mathrmS_1$ and $mathrmS_2$ are two circles of unit radius touching at point $P$; $T$ is a common tangent to $mathrmS_1$ and $mathrmS_2$ touching them at $X$ and $Y$; $mathrmC_1$ is the circle touching $mathrmS_1$, $mathrmS_2$, and $T$. $mathrmC_n$ is the circle touching $mathrmS_1$ , $mathrmS_2$, and $mathrmC_n-1$ for $n > 1.$ By computing diameters of $mathrmC_n$ prove that $$frac11cdot 2+frac12cdot 3+frac13cdot 4+ldots =1,.$$
sequences-and-series geometry convergence circle tangent-line
edited Aug 3 at 10:16
Batominovski
22.7k22776
asked Aug 2 at 14:50
Identicon
808
2
I believe you mix up $1$ and $I$ as well as $S_1$ and $C_n-1$ at some point.
– lattice
Aug 2 at 15:01
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up vote
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$mathrmS_1$ and $mathrmS_2$ are two circles of unit radius touching at point $P$; $T$ is a common tangent to $mathrmS_1$ and $mathrmS_2$ touching them at $X$ and $Y$; $mathrmC_1$ is the circle touching $mathrmS_1$, $mathrmS_2$, and $T$. $mathrmC_n$ is the circle touching $mathrmS_1$ , $mathrmS_2$, and $mathrmC_n-1$ for $n > 1.$ By computing diameters of $mathrmC_n$ prove that $$frac11cdot 2+frac12cdot 3+frac13cdot 4+ldots =1,.$$
sequences-and-series geometry convergence circle tangent-line
edited Aug 3 at 10:16
Batominovski
22.7k22776
asked Aug 2 at 14:50
Identicon
808
$mathrmS_1$ and $mathrmS_2$ are two circles of unit radius touching at point $P$; $T$ is a common tangent to $mathrmS_1$ and $mathrmS_2$ touching them at $X$ and $Y$; $mathrmC_1$ is the circle touching $mathrmS_1$, $mathrmS_2$, and $T$. $mathrmC_n$ is the circle touching $mathrmS_1$ , $mathrmS_2$, and $mathrmC_n-1$ for $n > 1.$ By computing diameters of $mathrmC_n$ prove that $$frac11cdot 2+frac12cdot 3+frac13cdot 4+ldots =1,.$$
sequences-and-series geometry convergence circle tangent-line
edited Aug 3 at 10:16
Batominovski
22.7k22776
asked Aug 2 at 14:50
Identicon
808
edited Aug 3 at 10:16
Batominovski
22.7k22776
edited Aug 3 at 10:16
Batominovski
22.7k22776
edited Aug 3 at 10:16
Batominovski
22.7k22776
22.7k22776
asked Aug 2 at 14:50
Identicon
808
asked Aug 2 at 14:50
Identicon
808
asked Aug 2 at 14:50
Identicon
808
808
2
I believe you mix up $1$ and $I$ as well as $S_1$ and $C_n-1$ at some point.
– lattice
Aug 2 at 15:01
add a comment |Â
2
I believe you mix up $1$ and $I$ as well as $S_1$ and $C_n-1$ at some point.
– lattice
Aug 2 at 15:01
2
2
I believe you mix up $1$ and $I$ as well as $S_1$ and $C_n-1$ at some point.
– lattice
Aug 2 at 15:01
I believe you mix up $1$ and $I$ as well as $S_1$ and $C_n-1$ at some point.
– lattice
Aug 2 at 15:01
add a comment |Â
2 Answers
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By using Descartes circle theorem we find that the curvature of circle $C_1$ is $4$, of $C_2$ is $12$, and in general, the curvature of circle $C_n$ is $2n(n+1)$.The diameter of circle $C_n$ is $frac1n(n+1)$. The $C_n$ circles are stacked on top of each other which implies the sum of all the diameters is the distance from the common tangent $T$ to the intersection point $P$. This is the radius of $S_1$ and $S_2$ which is $1$. Q.E.D.
The Wikipedia article gives
$ k_4 = K_pm(k_1, k_2, k_3) := k_1 + k_2 + k_3 pm 2sqrtk_1k_2 + k_2k_3 + k_3k_1, $ the formula to compute the curvature of one of the four curvatures of four mutually tangent circles in terms of the other three curvatures.
We begin with $ K_1 = 4 = K_+(1, 1, 0) $ since the tangent line has curvature $0$. Then $ K_2 = 12 = K_+1, 1, 4), quad K_3 = 24 = K_+1, 1, 12), $
and so on. You can check that $ K_+(1, 1, 2n(n+1)) = 2(n+1)(n+2). $ By definition, the curvature is the reciprocal of the radius, and thus the radius of circle $C_n$ is $ frac12n(n+1). $
answered Aug 2 at 22:55
Somos
10.9k1831
Sir, Can you pls explain how you used Descartes circle theorem to find the curvature. And how to find diameter from that. Please.
– Identicon
Aug 3 at 4:01
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Here is an alternative way to compute the diameter $d_n$ of $C_n$. Once you obtain $d_n$, you use the same argument as Somos presented to show that $sumlimits_n=1^infty,d_n=R$, where $R$ is the radius of $S_1$ and $S_2$ (here $R$ is set to be $1$).
Consider the inversion $i$ with center $P$ and radius $rho:=PX=PY=sqrt2R$. Then, $l_1:=i(S_1)$ and $l_2:=i(S_2)$ are two parallel lines perpendicular to $XY$ at $X$ and $Y$, respectively. The image $i(XY)$ of the line $XY$ is the circle $Gamma$ passing through $X$, $Y$, and $P$ (noting that $Gamma$ toughes both $l_1$ and $l_2$).
Now, let $gamma_n$ denote the image $i(C_n)$ of the circle $C_n$, for $n=1,2,3,ldots$. Write $PP_1$ for the diameter of $Gamma$ with one endpoint being $P$. It can be easily seen that there are points $P_2,P_3,ldots$ on the ray $PP_1$ such that $2R=PP_1=P_1P_2=P_2P_3=ldots$ and each $P_nP_n+1$ is a diameter of $gamma_n$. This means $PP_n=2Rn$. Now, write $Q_n$ for $i(P_n)$. Then,
$$PQ_n=fracrho^2PP_n=frac2R^22Rn=fracRn,.$$
That is, if $d_n$ is the diameter of $gamma_n$, then
$$d_n=Q_nQ_n+1=PQ_n-PQ_n+1=fracRn-fracRn+1=fracRn(n+1),.$$
answered Aug 3 at 9:06
Batominovski
22.7k22776
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
By using Descartes circle theorem we find that the curvature of circle $C_1$ is $4$, of $C_2$ is $12$, and in general, the curvature of circle $C_n$ is $2n(n+1)$.The diameter of circle $C_n$ is $frac1n(n+1)$. The $C_n$ circles are stacked on top of each other which implies the sum of all the diameters is the distance from the common tangent $T$ to the intersection point $P$. This is the radius of $S_1$ and $S_2$ which is $1$. Q.E.D.
The Wikipedia article gives
$ k_4 = K_pm(k_1, k_2, k_3) := k_1 + k_2 + k_3 pm 2sqrtk_1k_2 + k_2k_3 + k_3k_1, $ the formula to compute the curvature of one of the four curvatures of four mutually tangent circles in terms of the other three curvatures.
We begin with $ K_1 = 4 = K_+(1, 1, 0) $ since the tangent line has curvature $0$. Then $ K_2 = 12 = K_+1, 1, 4), quad K_3 = 24 = K_+1, 1, 12), $
and so on. You can check that $ K_+(1, 1, 2n(n+1)) = 2(n+1)(n+2). $ By definition, the curvature is the reciprocal of the radius, and thus the radius of circle $C_n$ is $ frac12n(n+1). $
answered Aug 2 at 22:55
Somos
10.9k1831
Sir, Can you pls explain how you used Descartes circle theorem to find the curvature. And how to find diameter from that. Please.
– Identicon
Aug 3 at 4:01
add a comment |Â
up vote
3
down vote
accepted
By using Descartes circle theorem we find that the curvature of circle $C_1$ is $4$, of $C_2$ is $12$, and in general, the curvature of circle $C_n$ is $2n(n+1)$.The diameter of circle $C_n$ is $frac1n(n+1)$. The $C_n$ circles are stacked on top of each other which implies the sum of all the diameters is the distance from the common tangent $T$ to the intersection point $P$. This is the radius of $S_1$ and $S_2$ which is $1$. Q.E.D.
The Wikipedia article gives
$ k_4 = K_pm(k_1, k_2, k_3) := k_1 + k_2 + k_3 pm 2sqrtk_1k_2 + k_2k_3 + k_3k_1, $ the formula to compute the curvature of one of the four curvatures of four mutually tangent circles in terms of the other three curvatures.
We begin with $ K_1 = 4 = K_+(1, 1, 0) $ since the tangent line has curvature $0$. Then $ K_2 = 12 = K_+1, 1, 4), quad K_3 = 24 = K_+1, 1, 12), $
and so on. You can check that $ K_+(1, 1, 2n(n+1)) = 2(n+1)(n+2). $ By definition, the curvature is the reciprocal of the radius, and thus the radius of circle $C_n$ is $ frac12n(n+1). $
answered Aug 2 at 22:55
Somos
10.9k1831
Sir, Can you pls explain how you used Descartes circle theorem to find the curvature. And how to find diameter from that. Please.
– Identicon
Aug 3 at 4:01
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
By using Descartes circle theorem we find that the curvature of circle $C_1$ is $4$, of $C_2$ is $12$, and in general, the curvature of circle $C_n$ is $2n(n+1)$.The diameter of circle $C_n$ is $frac1n(n+1)$. The $C_n$ circles are stacked on top of each other which implies the sum of all the diameters is the distance from the common tangent $T$ to the intersection point $P$. This is the radius of $S_1$ and $S_2$ which is $1$. Q.E.D.
The Wikipedia article gives
$ k_4 = K_pm(k_1, k_2, k_3) := k_1 + k_2 + k_3 pm 2sqrtk_1k_2 + k_2k_3 + k_3k_1, $ the formula to compute the curvature of one of the four curvatures of four mutually tangent circles in terms of the other three curvatures.
We begin with $ K_1 = 4 = K_+(1, 1, 0) $ since the tangent line has curvature $0$. Then $ K_2 = 12 = K_+1, 1, 4), quad K_3 = 24 = K_+1, 1, 12), $
and so on. You can check that $ K_+(1, 1, 2n(n+1)) = 2(n+1)(n+2). $ By definition, the curvature is the reciprocal of the radius, and thus the radius of circle $C_n$ is $ frac12n(n+1). $
answered Aug 2 at 22:55
Somos
10.9k1831
By using Descartes circle theorem we find that the curvature of circle $C_1$ is $4$, of $C_2$ is $12$, and in general, the curvature of circle $C_n$ is $2n(n+1)$.The diameter of circle $C_n$ is $frac1n(n+1)$. The $C_n$ circles are stacked on top of each other which implies the sum of all the diameters is the distance from the common tangent $T$ to the intersection point $P$. This is the radius of $S_1$ and $S_2$ which is $1$. Q.E.D.
The Wikipedia article gives
$ k_4 = K_pm(k_1, k_2, k_3) := k_1 + k_2 + k_3 pm 2sqrtk_1k_2 + k_2k_3 + k_3k_1, $ the formula to compute the curvature of one of the four curvatures of four mutually tangent circles in terms of the other three curvatures.
We begin with $ K_1 = 4 = K_+(1, 1, 0) $ since the tangent line has curvature $0$. Then $ K_2 = 12 = K_+1, 1, 4), quad K_3 = 24 = K_+1, 1, 12), $
and so on. You can check that $ K_+(1, 1, 2n(n+1)) = 2(n+1)(n+2). $ By definition, the curvature is the reciprocal of the radius, and thus the radius of circle $C_n$ is $ frac12n(n+1). $
answered Aug 2 at 22:55
Somos
10.9k1831
edited Aug 3 at 10:52
answered Aug 2 at 22:55
Somos
10.9k1831
answered Aug 2 at 22:55
Somos
10.9k1831
answered Aug 2 at 22:55
Somos
10.9k1831
10.9k1831
Sir, Can you pls explain how you used Descartes circle theorem to find the curvature. And how to find diameter from that. Please.
– Identicon
Aug 3 at 4:01
add a comment |Â
Sir, Can you pls explain how you used Descartes circle theorem to find the curvature. And how to find diameter from that. Please.
– Identicon
Aug 3 at 4:01
Sir, Can you pls explain how you used Descartes circle theorem to find the curvature. And how to find diameter from that. Please.
– Identicon
Aug 3 at 4:01
Sir, Can you pls explain how you used Descartes circle theorem to find the curvature. And how to find diameter from that. Please.
– Identicon
Aug 3 at 4:01
add a comment |Â
up vote
2
down vote
Here is an alternative way to compute the diameter $d_n$ of $C_n$. Once you obtain $d_n$, you use the same argument as Somos presented to show that $sumlimits_n=1^infty,d_n=R$, where $R$ is the radius of $S_1$ and $S_2$ (here $R$ is set to be $1$).
Consider the inversion $i$ with center $P$ and radius $rho:=PX=PY=sqrt2R$. Then, $l_1:=i(S_1)$ and $l_2:=i(S_2)$ are two parallel lines perpendicular to $XY$ at $X$ and $Y$, respectively. The image $i(XY)$ of the line $XY$ is the circle $Gamma$ passing through $X$, $Y$, and $P$ (noting that $Gamma$ toughes both $l_1$ and $l_2$).
Now, let $gamma_n$ denote the image $i(C_n)$ of the circle $C_n$, for $n=1,2,3,ldots$. Write $PP_1$ for the diameter of $Gamma$ with one endpoint being $P$. It can be easily seen that there are points $P_2,P_3,ldots$ on the ray $PP_1$ such that $2R=PP_1=P_1P_2=P_2P_3=ldots$ and each $P_nP_n+1$ is a diameter of $gamma_n$. This means $PP_n=2Rn$. Now, write $Q_n$ for $i(P_n)$. Then,
$$PQ_n=fracrho^2PP_n=frac2R^22Rn=fracRn,.$$
That is, if $d_n$ is the diameter of $gamma_n$, then
$$d_n=Q_nQ_n+1=PQ_n-PQ_n+1=fracRn-fracRn+1=fracRn(n+1),.$$
answered Aug 3 at 9:06
Batominovski
22.7k22776
add a comment |Â
up vote
2
down vote
Here is an alternative way to compute the diameter $d_n$ of $C_n$. Once you obtain $d_n$, you use the same argument as Somos presented to show that $sumlimits_n=1^infty,d_n=R$, where $R$ is the radius of $S_1$ and $S_2$ (here $R$ is set to be $1$).
Consider the inversion $i$ with center $P$ and radius $rho:=PX=PY=sqrt2R$. Then, $l_1:=i(S_1)$ and $l_2:=i(S_2)$ are two parallel lines perpendicular to $XY$ at $X$ and $Y$, respectively. The image $i(XY)$ of the line $XY$ is the circle $Gamma$ passing through $X$, $Y$, and $P$ (noting that $Gamma$ toughes both $l_1$ and $l_2$).
Now, let $gamma_n$ denote the image $i(C_n)$ of the circle $C_n$, for $n=1,2,3,ldots$. Write $PP_1$ for the diameter of $Gamma$ with one endpoint being $P$. It can be easily seen that there are points $P_2,P_3,ldots$ on the ray $PP_1$ such that $2R=PP_1=P_1P_2=P_2P_3=ldots$ and each $P_nP_n+1$ is a diameter of $gamma_n$. This means $PP_n=2Rn$. Now, write $Q_n$ for $i(P_n)$. Then,
$$PQ_n=fracrho^2PP_n=frac2R^22Rn=fracRn,.$$
That is, if $d_n$ is the diameter of $gamma_n$, then
$$d_n=Q_nQ_n+1=PQ_n-PQ_n+1=fracRn-fracRn+1=fracRn(n+1),.$$
answered Aug 3 at 9:06
Batominovski
22.7k22776
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Here is an alternative way to compute the diameter $d_n$ of $C_n$. Once you obtain $d_n$, you use the same argument as Somos presented to show that $sumlimits_n=1^infty,d_n=R$, where $R$ is the radius of $S_1$ and $S_2$ (here $R$ is set to be $1$).
Consider the inversion $i$ with center $P$ and radius $rho:=PX=PY=sqrt2R$. Then, $l_1:=i(S_1)$ and $l_2:=i(S_2)$ are two parallel lines perpendicular to $XY$ at $X$ and $Y$, respectively. The image $i(XY)$ of the line $XY$ is the circle $Gamma$ passing through $X$, $Y$, and $P$ (noting that $Gamma$ toughes both $l_1$ and $l_2$).
Now, let $gamma_n$ denote the image $i(C_n)$ of the circle $C_n$, for $n=1,2,3,ldots$. Write $PP_1$ for the diameter of $Gamma$ with one endpoint being $P$. It can be easily seen that there are points $P_2,P_3,ldots$ on the ray $PP_1$ such that $2R=PP_1=P_1P_2=P_2P_3=ldots$ and each $P_nP_n+1$ is a diameter of $gamma_n$. This means $PP_n=2Rn$. Now, write $Q_n$ for $i(P_n)$. Then,
$$PQ_n=fracrho^2PP_n=frac2R^22Rn=fracRn,.$$
That is, if $d_n$ is the diameter of $gamma_n$, then
$$d_n=Q_nQ_n+1=PQ_n-PQ_n+1=fracRn-fracRn+1=fracRn(n+1),.$$
answered Aug 3 at 9:06
Batominovski
22.7k22776
Here is an alternative way to compute the diameter $d_n$ of $C_n$. Once you obtain $d_n$, you use the same argument as Somos presented to show that $sumlimits_n=1^infty,d_n=R$, where $R$ is the radius of $S_1$ and $S_2$ (here $R$ is set to be $1$).
Consider the inversion $i$ with center $P$ and radius $rho:=PX=PY=sqrt2R$. Then, $l_1:=i(S_1)$ and $l_2:=i(S_2)$ are two parallel lines perpendicular to $XY$ at $X$ and $Y$, respectively. The image $i(XY)$ of the line $XY$ is the circle $Gamma$ passing through $X$, $Y$, and $P$ (noting that $Gamma$ toughes both $l_1$ and $l_2$).
Now, let $gamma_n$ denote the image $i(C_n)$ of the circle $C_n$, for $n=1,2,3,ldots$. Write $PP_1$ for the diameter of $Gamma$ with one endpoint being $P$. It can be easily seen that there are points $P_2,P_3,ldots$ on the ray $PP_1$ such that $2R=PP_1=P_1P_2=P_2P_3=ldots$ and each $P_nP_n+1$ is a diameter of $gamma_n$. This means $PP_n=2Rn$. Now, write $Q_n$ for $i(P_n)$. Then,
$$PQ_n=fracrho^2PP_n=frac2R^22Rn=fracRn,.$$
That is, if $d_n$ is the diameter of $gamma_n$, then
$$d_n=Q_nQ_n+1=PQ_n-PQ_n+1=fracRn-fracRn+1=fracRn(n+1),.$$
answered Aug 3 at 9:06
Batominovski
22.7k22776
answered Aug 3 at 9:06
Batominovski
22.7k22776
answered Aug 3 at 9:06
Batominovski
22.7k22776
answered Aug 3 at 9:06
Batominovski
22.7k22776
22.7k22776
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2
I believe you mix up $1$ and $I$ as well as $S_1$ and $C_n-1$ at some point.
– lattice
Aug 2 at 15:01