Infinite Series $sumlimits_{n=0}^{infty}arctan(frac{1}{F_{2n+1}})$












6














How can I find the value of the following sum?
$$sum_{n=0}^{infty}arctan(frac{1}{F_{2n+1}})$$
$F_n$ is the Fibonacci number.($F_1=F_2=1$)






calculus sequences-and-series analysis







share|cite|improve this question
























  • Just to be sure about the indices, $F_1 = F_2 = 1$?
    – Dan Shved
    Nov 5 '13 at 9:55










  • Well, it clearly is $frac{pi}{2}$. Just need to actually prove it )
    – Dan Shved
    Nov 5 '13 at 10:03
















6














How can I find the value of the following sum?
$$sum_{n=0}^{infty}arctan(frac{1}{F_{2n+1}})$$
$F_n$ is the Fibonacci number.($F_1=F_2=1$)






calculus sequences-and-series analysis







share|cite|improve this question
























  • Just to be sure about the indices, $F_1 = F_2 = 1$?
    – Dan Shved
    Nov 5 '13 at 9:55










  • Well, it clearly is $frac{pi}{2}$. Just need to actually prove it )
    – Dan Shved
    Nov 5 '13 at 10:03














6












6








6


8





How can I find the value of the following sum?
$$sum_{n=0}^{infty}arctan(frac{1}{F_{2n+1}})$$
$F_n$ is the Fibonacci number.($F_1=F_2=1$)






calculus sequences-and-series analysis







share|cite|improve this question















How can I find the value of the following sum?
$$sum_{n=0}^{infty}arctan(frac{1}{F_{2n+1}})$$
$F_n$ is the Fibonacci number.($F_1=F_2=1$)






calculus sequences-and-series analysis




calculus sequences-and-series analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 11:55









user91500

3,590946105




3,590946105










asked Nov 5 '13 at 9:49







user95733



















  • Just to be sure about the indices, $F_1 = F_2 = 1$?
    – Dan Shved
    Nov 5 '13 at 9:55










  • Well, it clearly is $frac{pi}{2}$. Just need to actually prove it )
    – Dan Shved
    Nov 5 '13 at 10:03


















  • Just to be sure about the indices, $F_1 = F_2 = 1$?
    – Dan Shved
    Nov 5 '13 at 9:55










  • Well, it clearly is $frac{pi}{2}$. Just need to actually prove it )
    – Dan Shved
    Nov 5 '13 at 10:03
















Just to be sure about the indices, $F_1 = F_2 = 1$?
– Dan Shved
Nov 5 '13 at 9:55




Just to be sure about the indices, $F_1 = F_2 = 1$?
– Dan Shved
Nov 5 '13 at 9:55












Well, it clearly is $frac{pi}{2}$. Just need to actually prove it )
– Dan Shved
Nov 5 '13 at 10:03




Well, it clearly is $frac{pi}{2}$. Just need to actually prove it )
– Dan Shved
Nov 5 '13 at 10:03










2 Answers
2






active

oldest

votes


















8














OK, let us denote by $a_n$ this complex number: $$a_n = (F_1 + i)(F_3 + i)ldots(F_{2n-1} + i).$$ I claim that for every $n geq 1$ we have $a_n = C cdot (1 + F_{2n} i)$, where $C$ is a positive real number (that depends on $n$).



Let us prove this by induction. For $n = 1$ we have $$a_1 = F_1 + i = 1 + i = 1 + F_2 i.$$



Now the transition. Suppose we have proved that $a_n = C(1 + F_{2n}i)$, where $C$ is a positive real. Then
$$
a_{n+1} = a_n (F_{2n+1} + i) = C (1 + F_{2n}i)(F_{2n+1} + i) = C(F_{2n+1}-F_{2n} + icdot(F_{2n} F_{2n+1} + 1)).
$$
Now, from the equalities on wikipedia it's easy to derive that $F_{2n}F_{2n+1} + 1 = F_{2n-1}F_{2n+2}$. Then we have
$$
a_{n+1} = CF_{2n-1}(1 + F_{2n+2}i).
$$
$CF_{2n-1}$ is a positive real number, so this completes the proof.



Now we are ready to prove that your infinite sum is equal to $pi/2$. If we look at the partial sum, we easily find that
$$
sum_{n=0}^{k}arctan(frac{1}{F_{2n+1}}) = sum_{n=0}^{k}arg (F_{2n+1} + i) = arg a_{k+1} = arctan(F_{2k + 2}).
$$
As $k$ tends to $+infty$, $F_{2k+2}$ also tends to $+infty$, and its $arctan$ tends to $pi/2$. So the answer is $pi/2$.






share|cite|improve this answer























  • +1 Nice use of complex numbers! I was so sold on sticking to Fibonacci identities, and the basic identities of tangent.
    – Jyrki Lahtonen
    Nov 5 '13 at 10:51





















9














We can prove by induction that
$$
sum_{n=0}^karctanfrac1{F_{2n+1}}=arctan F_{2k+2}.
$$
The base case $k=0$ is immediate, because $F_1=F_2=1$.



OTOH by induction hypothesis
$$
sum_{n=0}^{k+1}arctanfrac1{F_{2n+1}}=arctan F_{2k+2}+arctanfrac1{F_{2k+3}}.
$$
It follows from the formula for the tangent of the sum of two angles (careful about the overflow!)
$$
arctan x+arctan yequiv arctanfrac{x+y}{1-xy}pmodpi.
$$
Here $x=F_{2k+2}$, $y=1/F_{2k+3}$, and thus
$$
begin{aligned}
frac{x+y}{1-xy}&=frac{F_{2k+2}F_{2k+3}+1}{F_{2k+3}-F_{2k+2}}\
&=frac{F_{2k+2}F_{2k+3}+1}{F_{2k+1}}=F_{2k+4}
end{aligned}
$$
by the identity $F_{2k+4}F_{2k+1}=1+F_{2k+2}F_{2k+3}$ completing our induction step (see the link given by Dan Shved or prove this identity by induction).



As $F_ntoinfty$ it follows that the limit is $pi/2$.






share|cite|improve this answer























  • A cute problem. Thanks to whoever came up with this. I want to use this as an extra HW problem some day.
    – Jyrki Lahtonen
    Nov 5 '13 at 10:52










  • I kind of cheated in my answer by not mentioning the possibility of an overflow. Thanks for bringing that up, I guess :)
    – Dan Shved
    Nov 5 '13 at 10:59










  • Well, I didn't really deal with it either. Adding one $arctan$ at a time, as in an induction, sorta prevents the overflows I think :-/
    – Jyrki Lahtonen
    Nov 5 '13 at 11:02











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









8














OK, let us denote by $a_n$ this complex number: $$a_n = (F_1 + i)(F_3 + i)ldots(F_{2n-1} + i).$$ I claim that for every $n geq 1$ we have $a_n = C cdot (1 + F_{2n} i)$, where $C$ is a positive real number (that depends on $n$).



Let us prove this by induction. For $n = 1$ we have $$a_1 = F_1 + i = 1 + i = 1 + F_2 i.$$



Now the transition. Suppose we have proved that $a_n = C(1 + F_{2n}i)$, where $C$ is a positive real. Then
$$
a_{n+1} = a_n (F_{2n+1} + i) = C (1 + F_{2n}i)(F_{2n+1} + i) = C(F_{2n+1}-F_{2n} + icdot(F_{2n} F_{2n+1} + 1)).
$$
Now, from the equalities on wikipedia it's easy to derive that $F_{2n}F_{2n+1} + 1 = F_{2n-1}F_{2n+2}$. Then we have
$$
a_{n+1} = CF_{2n-1}(1 + F_{2n+2}i).
$$
$CF_{2n-1}$ is a positive real number, so this completes the proof.



Now we are ready to prove that your infinite sum is equal to $pi/2$. If we look at the partial sum, we easily find that
$$
sum_{n=0}^{k}arctan(frac{1}{F_{2n+1}}) = sum_{n=0}^{k}arg (F_{2n+1} + i) = arg a_{k+1} = arctan(F_{2k + 2}).
$$
As $k$ tends to $+infty$, $F_{2k+2}$ also tends to $+infty$, and its $arctan$ tends to $pi/2$. So the answer is $pi/2$.






share|cite|improve this answer























  • +1 Nice use of complex numbers! I was so sold on sticking to Fibonacci identities, and the basic identities of tangent.
    – Jyrki Lahtonen
    Nov 5 '13 at 10:51


















8














OK, let us denote by $a_n$ this complex number: $$a_n = (F_1 + i)(F_3 + i)ldots(F_{2n-1} + i).$$ I claim that for every $n geq 1$ we have $a_n = C cdot (1 + F_{2n} i)$, where $C$ is a positive real number (that depends on $n$).



Let us prove this by induction. For $n = 1$ we have $$a_1 = F_1 + i = 1 + i = 1 + F_2 i.$$



Now the transition. Suppose we have proved that $a_n = C(1 + F_{2n}i)$, where $C$ is a positive real. Then
$$
a_{n+1} = a_n (F_{2n+1} + i) = C (1 + F_{2n}i)(F_{2n+1} + i) = C(F_{2n+1}-F_{2n} + icdot(F_{2n} F_{2n+1} + 1)).
$$
Now, from the equalities on wikipedia it's easy to derive that $F_{2n}F_{2n+1} + 1 = F_{2n-1}F_{2n+2}$. Then we have
$$
a_{n+1} = CF_{2n-1}(1 + F_{2n+2}i).
$$
$CF_{2n-1}$ is a positive real number, so this completes the proof.



Now we are ready to prove that your infinite sum is equal to $pi/2$. If we look at the partial sum, we easily find that
$$
sum_{n=0}^{k}arctan(frac{1}{F_{2n+1}}) = sum_{n=0}^{k}arg (F_{2n+1} + i) = arg a_{k+1} = arctan(F_{2k + 2}).
$$
As $k$ tends to $+infty$, $F_{2k+2}$ also tends to $+infty$, and its $arctan$ tends to $pi/2$. So the answer is $pi/2$.






share|cite|improve this answer























  • +1 Nice use of complex numbers! I was so sold on sticking to Fibonacci identities, and the basic identities of tangent.
    – Jyrki Lahtonen
    Nov 5 '13 at 10:51
















8












8








8






OK, let us denote by $a_n$ this complex number: $$a_n = (F_1 + i)(F_3 + i)ldots(F_{2n-1} + i).$$ I claim that for every $n geq 1$ we have $a_n = C cdot (1 + F_{2n} i)$, where $C$ is a positive real number (that depends on $n$).



Let us prove this by induction. For $n = 1$ we have $$a_1 = F_1 + i = 1 + i = 1 + F_2 i.$$



Now the transition. Suppose we have proved that $a_n = C(1 + F_{2n}i)$, where $C$ is a positive real. Then
$$
a_{n+1} = a_n (F_{2n+1} + i) = C (1 + F_{2n}i)(F_{2n+1} + i) = C(F_{2n+1}-F_{2n} + icdot(F_{2n} F_{2n+1} + 1)).
$$
Now, from the equalities on wikipedia it's easy to derive that $F_{2n}F_{2n+1} + 1 = F_{2n-1}F_{2n+2}$. Then we have
$$
a_{n+1} = CF_{2n-1}(1 + F_{2n+2}i).
$$
$CF_{2n-1}$ is a positive real number, so this completes the proof.



Now we are ready to prove that your infinite sum is equal to $pi/2$. If we look at the partial sum, we easily find that
$$
sum_{n=0}^{k}arctan(frac{1}{F_{2n+1}}) = sum_{n=0}^{k}arg (F_{2n+1} + i) = arg a_{k+1} = arctan(F_{2k + 2}).
$$
As $k$ tends to $+infty$, $F_{2k+2}$ also tends to $+infty$, and its $arctan$ tends to $pi/2$. So the answer is $pi/2$.






share|cite|improve this answer














OK, let us denote by $a_n$ this complex number: $$a_n = (F_1 + i)(F_3 + i)ldots(F_{2n-1} + i).$$ I claim that for every $n geq 1$ we have $a_n = C cdot (1 + F_{2n} i)$, where $C$ is a positive real number (that depends on $n$).



Let us prove this by induction. For $n = 1$ we have $$a_1 = F_1 + i = 1 + i = 1 + F_2 i.$$



Now the transition. Suppose we have proved that $a_n = C(1 + F_{2n}i)$, where $C$ is a positive real. Then
$$
a_{n+1} = a_n (F_{2n+1} + i) = C (1 + F_{2n}i)(F_{2n+1} + i) = C(F_{2n+1}-F_{2n} + icdot(F_{2n} F_{2n+1} + 1)).
$$
Now, from the equalities on wikipedia it's easy to derive that $F_{2n}F_{2n+1} + 1 = F_{2n-1}F_{2n+2}$. Then we have
$$
a_{n+1} = CF_{2n-1}(1 + F_{2n+2}i).
$$
$CF_{2n-1}$ is a positive real number, so this completes the proof.



Now we are ready to prove that your infinite sum is equal to $pi/2$. If we look at the partial sum, we easily find that
$$
sum_{n=0}^{k}arctan(frac{1}{F_{2n+1}}) = sum_{n=0}^{k}arg (F_{2n+1} + i) = arg a_{k+1} = arctan(F_{2k + 2}).
$$
As $k$ tends to $+infty$, $F_{2k+2}$ also tends to $+infty$, and its $arctan$ tends to $pi/2$. So the answer is $pi/2$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 5 '13 at 10:38

























answered Nov 5 '13 at 10:22









Dan Shved

12.6k2045




12.6k2045












  • +1 Nice use of complex numbers! I was so sold on sticking to Fibonacci identities, and the basic identities of tangent.
    – Jyrki Lahtonen
    Nov 5 '13 at 10:51




















  • +1 Nice use of complex numbers! I was so sold on sticking to Fibonacci identities, and the basic identities of tangent.
    – Jyrki Lahtonen
    Nov 5 '13 at 10:51


















+1 Nice use of complex numbers! I was so sold on sticking to Fibonacci identities, and the basic identities of tangent.
– Jyrki Lahtonen
Nov 5 '13 at 10:51






+1 Nice use of complex numbers! I was so sold on sticking to Fibonacci identities, and the basic identities of tangent.
– Jyrki Lahtonen
Nov 5 '13 at 10:51













9














We can prove by induction that
$$
sum_{n=0}^karctanfrac1{F_{2n+1}}=arctan F_{2k+2}.
$$
The base case $k=0$ is immediate, because $F_1=F_2=1$.



OTOH by induction hypothesis
$$
sum_{n=0}^{k+1}arctanfrac1{F_{2n+1}}=arctan F_{2k+2}+arctanfrac1{F_{2k+3}}.
$$
It follows from the formula for the tangent of the sum of two angles (careful about the overflow!)
$$
arctan x+arctan yequiv arctanfrac{x+y}{1-xy}pmodpi.
$$
Here $x=F_{2k+2}$, $y=1/F_{2k+3}$, and thus
$$
begin{aligned}
frac{x+y}{1-xy}&=frac{F_{2k+2}F_{2k+3}+1}{F_{2k+3}-F_{2k+2}}\
&=frac{F_{2k+2}F_{2k+3}+1}{F_{2k+1}}=F_{2k+4}
end{aligned}
$$
by the identity $F_{2k+4}F_{2k+1}=1+F_{2k+2}F_{2k+3}$ completing our induction step (see the link given by Dan Shved or prove this identity by induction).



As $F_ntoinfty$ it follows that the limit is $pi/2$.






share|cite|improve this answer























  • A cute problem. Thanks to whoever came up with this. I want to use this as an extra HW problem some day.
    – Jyrki Lahtonen
    Nov 5 '13 at 10:52










  • I kind of cheated in my answer by not mentioning the possibility of an overflow. Thanks for bringing that up, I guess :)
    – Dan Shved
    Nov 5 '13 at 10:59










  • Well, I didn't really deal with it either. Adding one $arctan$ at a time, as in an induction, sorta prevents the overflows I think :-/
    – Jyrki Lahtonen
    Nov 5 '13 at 11:02
















9














We can prove by induction that
$$
sum_{n=0}^karctanfrac1{F_{2n+1}}=arctan F_{2k+2}.
$$
The base case $k=0$ is immediate, because $F_1=F_2=1$.



OTOH by induction hypothesis
$$
sum_{n=0}^{k+1}arctanfrac1{F_{2n+1}}=arctan F_{2k+2}+arctanfrac1{F_{2k+3}}.
$$
It follows from the formula for the tangent of the sum of two angles (careful about the overflow!)
$$
arctan x+arctan yequiv arctanfrac{x+y}{1-xy}pmodpi.
$$
Here $x=F_{2k+2}$, $y=1/F_{2k+3}$, and thus
$$
begin{aligned}
frac{x+y}{1-xy}&=frac{F_{2k+2}F_{2k+3}+1}{F_{2k+3}-F_{2k+2}}\
&=frac{F_{2k+2}F_{2k+3}+1}{F_{2k+1}}=F_{2k+4}
end{aligned}
$$
by the identity $F_{2k+4}F_{2k+1}=1+F_{2k+2}F_{2k+3}$ completing our induction step (see the link given by Dan Shved or prove this identity by induction).



As $F_ntoinfty$ it follows that the limit is $pi/2$.






share|cite|improve this answer























  • A cute problem. Thanks to whoever came up with this. I want to use this as an extra HW problem some day.
    – Jyrki Lahtonen
    Nov 5 '13 at 10:52










  • I kind of cheated in my answer by not mentioning the possibility of an overflow. Thanks for bringing that up, I guess :)
    – Dan Shved
    Nov 5 '13 at 10:59










  • Well, I didn't really deal with it either. Adding one $arctan$ at a time, as in an induction, sorta prevents the overflows I think :-/
    – Jyrki Lahtonen
    Nov 5 '13 at 11:02














9












9








9






We can prove by induction that
$$
sum_{n=0}^karctanfrac1{F_{2n+1}}=arctan F_{2k+2}.
$$
The base case $k=0$ is immediate, because $F_1=F_2=1$.



OTOH by induction hypothesis
$$
sum_{n=0}^{k+1}arctanfrac1{F_{2n+1}}=arctan F_{2k+2}+arctanfrac1{F_{2k+3}}.
$$
It follows from the formula for the tangent of the sum of two angles (careful about the overflow!)
$$
arctan x+arctan yequiv arctanfrac{x+y}{1-xy}pmodpi.
$$
Here $x=F_{2k+2}$, $y=1/F_{2k+3}$, and thus
$$
begin{aligned}
frac{x+y}{1-xy}&=frac{F_{2k+2}F_{2k+3}+1}{F_{2k+3}-F_{2k+2}}\
&=frac{F_{2k+2}F_{2k+3}+1}{F_{2k+1}}=F_{2k+4}
end{aligned}
$$
by the identity $F_{2k+4}F_{2k+1}=1+F_{2k+2}F_{2k+3}$ completing our induction step (see the link given by Dan Shved or prove this identity by induction).



As $F_ntoinfty$ it follows that the limit is $pi/2$.






share|cite|improve this answer














We can prove by induction that
$$
sum_{n=0}^karctanfrac1{F_{2n+1}}=arctan F_{2k+2}.
$$
The base case $k=0$ is immediate, because $F_1=F_2=1$.



OTOH by induction hypothesis
$$
sum_{n=0}^{k+1}arctanfrac1{F_{2n+1}}=arctan F_{2k+2}+arctanfrac1{F_{2k+3}}.
$$
It follows from the formula for the tangent of the sum of two angles (careful about the overflow!)
$$
arctan x+arctan yequiv arctanfrac{x+y}{1-xy}pmodpi.
$$
Here $x=F_{2k+2}$, $y=1/F_{2k+3}$, and thus
$$
begin{aligned}
frac{x+y}{1-xy}&=frac{F_{2k+2}F_{2k+3}+1}{F_{2k+3}-F_{2k+2}}\
&=frac{F_{2k+2}F_{2k+3}+1}{F_{2k+1}}=F_{2k+4}
end{aligned}
$$
by the identity $F_{2k+4}F_{2k+1}=1+F_{2k+2}F_{2k+3}$ completing our induction step (see the link given by Dan Shved or prove this identity by induction).



As $F_ntoinfty$ it follows that the limit is $pi/2$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 5 '13 at 11:01


























community wiki





2 revs
Jyrki Lahtonen













  • A cute problem. Thanks to whoever came up with this. I want to use this as an extra HW problem some day.
    – Jyrki Lahtonen
    Nov 5 '13 at 10:52










  • I kind of cheated in my answer by not mentioning the possibility of an overflow. Thanks for bringing that up, I guess :)
    – Dan Shved
    Nov 5 '13 at 10:59










  • Well, I didn't really deal with it either. Adding one $arctan$ at a time, as in an induction, sorta prevents the overflows I think :-/
    – Jyrki Lahtonen
    Nov 5 '13 at 11:02


















  • A cute problem. Thanks to whoever came up with this. I want to use this as an extra HW problem some day.
    – Jyrki Lahtonen
    Nov 5 '13 at 10:52










  • I kind of cheated in my answer by not mentioning the possibility of an overflow. Thanks for bringing that up, I guess :)
    – Dan Shved
    Nov 5 '13 at 10:59










  • Well, I didn't really deal with it either. Adding one $arctan$ at a time, as in an induction, sorta prevents the overflows I think :-/
    – Jyrki Lahtonen
    Nov 5 '13 at 11:02
















A cute problem. Thanks to whoever came up with this. I want to use this as an extra HW problem some day.
– Jyrki Lahtonen
Nov 5 '13 at 10:52




A cute problem. Thanks to whoever came up with this. I want to use this as an extra HW problem some day.
– Jyrki Lahtonen
Nov 5 '13 at 10:52












I kind of cheated in my answer by not mentioning the possibility of an overflow. Thanks for bringing that up, I guess :)
– Dan Shved
Nov 5 '13 at 10:59




I kind of cheated in my answer by not mentioning the possibility of an overflow. Thanks for bringing that up, I guess :)
– Dan Shved
Nov 5 '13 at 10:59












Well, I didn't really deal with it either. Adding one $arctan$ at a time, as in an induction, sorta prevents the overflows I think :-/
– Jyrki Lahtonen
Nov 5 '13 at 11:02




Well, I didn't really deal with it either. Adding one $arctan$ at a time, as in an induction, sorta prevents the overflows I think :-/
– Jyrki Lahtonen
Nov 5 '13 at 11:02


















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