A sum of Fibonacci numbers












6












$begingroup$


Let $F_n$ be the Fibonacci numbers. I would like to prove this really messy identity:
$$sum_{n=0}^{infty}(-1)^n(n+1)^2frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n+2choose 1}{2n+2 choose 2}{2n choose n}}=F_{k}tag1$$
which simplifies to:
$$sum_{n=0}^{infty}(-1)^nfrac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n choose n}(4n+2)}=F_{k}$$



How can I evaluate the following sums to show this identity (1) is correct?



$$sum_{n=0}^{infty}(-1)^nfrac{n^2}{{2n choose n}(2n+1)}=A$$



$$sum_{n=0}^{infty}(-1)^nfrac{n}{{2n choose n}(2n+1)}=B$$



$$sum_{n=0}^{infty}(-1)^nfrac{1}{{2n choose n}(2n+1)}=C$$






sequences-and-series fibonacci-numbers







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  • 6




    $begingroup$
    Wow, how did you come across this monstrosity?
    $endgroup$
    – YiFan
    3 hours ago
















6












$begingroup$


Let $F_n$ be the Fibonacci numbers. I would like to prove this really messy identity:
$$sum_{n=0}^{infty}(-1)^n(n+1)^2frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n+2choose 1}{2n+2 choose 2}{2n choose n}}=F_{k}tag1$$
which simplifies to:
$$sum_{n=0}^{infty}(-1)^nfrac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n choose n}(4n+2)}=F_{k}$$



How can I evaluate the following sums to show this identity (1) is correct?



$$sum_{n=0}^{infty}(-1)^nfrac{n^2}{{2n choose n}(2n+1)}=A$$



$$sum_{n=0}^{infty}(-1)^nfrac{n}{{2n choose n}(2n+1)}=B$$



$$sum_{n=0}^{infty}(-1)^nfrac{1}{{2n choose n}(2n+1)}=C$$






sequences-and-series fibonacci-numbers







share|cite|improve this question









New contributor




coffeee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 6




    $begingroup$
    Wow, how did you come across this monstrosity?
    $endgroup$
    – YiFan
    3 hours ago














6












6








6





$begingroup$


Let $F_n$ be the Fibonacci numbers. I would like to prove this really messy identity:
$$sum_{n=0}^{infty}(-1)^n(n+1)^2frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n+2choose 1}{2n+2 choose 2}{2n choose n}}=F_{k}tag1$$
which simplifies to:
$$sum_{n=0}^{infty}(-1)^nfrac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n choose n}(4n+2)}=F_{k}$$



How can I evaluate the following sums to show this identity (1) is correct?



$$sum_{n=0}^{infty}(-1)^nfrac{n^2}{{2n choose n}(2n+1)}=A$$



$$sum_{n=0}^{infty}(-1)^nfrac{n}{{2n choose n}(2n+1)}=B$$



$$sum_{n=0}^{infty}(-1)^nfrac{1}{{2n choose n}(2n+1)}=C$$






sequences-and-series fibonacci-numbers







share|cite|improve this question









New contributor




coffeee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Let $F_n$ be the Fibonacci numbers. I would like to prove this really messy identity:
$$sum_{n=0}^{infty}(-1)^n(n+1)^2frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n+2choose 1}{2n+2 choose 2}{2n choose n}}=F_{k}tag1$$
which simplifies to:
$$sum_{n=0}^{infty}(-1)^nfrac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n choose n}(4n+2)}=F_{k}$$



How can I evaluate the following sums to show this identity (1) is correct?



$$sum_{n=0}^{infty}(-1)^nfrac{n^2}{{2n choose n}(2n+1)}=A$$



$$sum_{n=0}^{infty}(-1)^nfrac{n}{{2n choose n}(2n+1)}=B$$



$$sum_{n=0}^{infty}(-1)^nfrac{1}{{2n choose n}(2n+1)}=C$$






sequences-and-series fibonacci-numbers




sequences-and-series fibonacci-numbers






share|cite|improve this question









New contributor




coffeee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




coffeee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









J. W. Tanner

1,308114




1,308114






New contributor




coffeee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 3 hours ago









coffeeecoffeee

311




311




New contributor




coffeee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





coffeee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






coffeee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 6




    $begingroup$
    Wow, how did you come across this monstrosity?
    $endgroup$
    – YiFan
    3 hours ago














  • 6




    $begingroup$
    Wow, how did you come across this monstrosity?
    $endgroup$
    – YiFan
    3 hours ago








6




6




$begingroup$
Wow, how did you come across this monstrosity?
$endgroup$
– YiFan
3 hours ago




$begingroup$
Wow, how did you come across this monstrosity?
$endgroup$
– YiFan
3 hours ago










1 Answer
1






active

oldest

votes


















6












$begingroup$

If we use Wolfram for these sums, with approximate values
begin{align*}
C approx 0.86082, qquad B approx -0.11649, qquad A approx -0.07897
end{align*}
From there, you can collect coefficients of $F_r$ to prove your identity. A more concrete way to derive exact expressions is through generating functions. Define
begin{align*}
g(x) = sum_{n=0}^{infty}frac{(-1)^n}{binom{2n}{n}(2n+1)}x^n
end{align*}
which you can show equals
begin{align*}
g(x) = 4 frac{sinh^{-1}(sqrt{x}/2)}{sqrt{x(x+4)}}
end{align*}
If we define $L = x frac{d}{dx}$ as the operator which first takes the derivative w.r.t $x$, then multiplies by $x$, we get that
begin{align*}
C = (L^0 g)(1), qquad B = (L^1 g)(1), qquad A = (L^2 g)(1)
end{align*}
or
begin{align*}
C = 4 frac{sinh^{-1}(1/2)}{sqrt{5}}, qquad B = frac{2}{5} - frac{12 sinh^{-1}(1/2)}{5sqrt{5}}, qquad A = frac{4}{125}(7sqrt{5}sinh^{-1}(1/2) - 10)
end{align*}






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Awesome answer! How did you come up with the generating functions and the corresponding closed form solution? Can you propose any resource for this?
    $endgroup$
    – MrYouMath
    2 hours ago






  • 1




    $begingroup$
    @MrYouMath: If you write $binom{2n}{n}=binom{2n}{2n-n}$, then the generating function $g(x)$ is the convolution of coefficients $a_n:=binom{2n}{n}^{-1}$ and $b_n:=(2n+1)^{-1}$, which is equivalent to the product of their individual generating functions.
    $endgroup$
    – Alex R.
    1 hour ago












  • $begingroup$
    Hi @MrYouMath, Alex's method is perhaps the best way to discover the generating function. Sometimes, you can even recognize the coefficients to be similar to existing Taylor expansions. In this case, the $a_n b_n$ collectively has similar form to the coefficients in the Taylor series expansion of $sinh^{-1}(x)$. From there, you can apply formal power series operations to try and and match the coefficients. The standard reference is generatingfunctionology by Wilf. Stanley's Enumerative Combinatorics also has an excellent presentation on generating functions / formal power series.
    $endgroup$
    – Tom Chen
    44 mins ago











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1 Answer
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1 Answer
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active

oldest

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active

oldest

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6












$begingroup$

If we use Wolfram for these sums, with approximate values
begin{align*}
C approx 0.86082, qquad B approx -0.11649, qquad A approx -0.07897
end{align*}
From there, you can collect coefficients of $F_r$ to prove your identity. A more concrete way to derive exact expressions is through generating functions. Define
begin{align*}
g(x) = sum_{n=0}^{infty}frac{(-1)^n}{binom{2n}{n}(2n+1)}x^n
end{align*}
which you can show equals
begin{align*}
g(x) = 4 frac{sinh^{-1}(sqrt{x}/2)}{sqrt{x(x+4)}}
end{align*}
If we define $L = x frac{d}{dx}$ as the operator which first takes the derivative w.r.t $x$, then multiplies by $x$, we get that
begin{align*}
C = (L^0 g)(1), qquad B = (L^1 g)(1), qquad A = (L^2 g)(1)
end{align*}
or
begin{align*}
C = 4 frac{sinh^{-1}(1/2)}{sqrt{5}}, qquad B = frac{2}{5} - frac{12 sinh^{-1}(1/2)}{5sqrt{5}}, qquad A = frac{4}{125}(7sqrt{5}sinh^{-1}(1/2) - 10)
end{align*}






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Awesome answer! How did you come up with the generating functions and the corresponding closed form solution? Can you propose any resource for this?
    $endgroup$
    – MrYouMath
    2 hours ago






  • 1




    $begingroup$
    @MrYouMath: If you write $binom{2n}{n}=binom{2n}{2n-n}$, then the generating function $g(x)$ is the convolution of coefficients $a_n:=binom{2n}{n}^{-1}$ and $b_n:=(2n+1)^{-1}$, which is equivalent to the product of their individual generating functions.
    $endgroup$
    – Alex R.
    1 hour ago












  • $begingroup$
    Hi @MrYouMath, Alex's method is perhaps the best way to discover the generating function. Sometimes, you can even recognize the coefficients to be similar to existing Taylor expansions. In this case, the $a_n b_n$ collectively has similar form to the coefficients in the Taylor series expansion of $sinh^{-1}(x)$. From there, you can apply formal power series operations to try and and match the coefficients. The standard reference is generatingfunctionology by Wilf. Stanley's Enumerative Combinatorics also has an excellent presentation on generating functions / formal power series.
    $endgroup$
    – Tom Chen
    44 mins ago
















6












$begingroup$

If we use Wolfram for these sums, with approximate values
begin{align*}
C approx 0.86082, qquad B approx -0.11649, qquad A approx -0.07897
end{align*}
From there, you can collect coefficients of $F_r$ to prove your identity. A more concrete way to derive exact expressions is through generating functions. Define
begin{align*}
g(x) = sum_{n=0}^{infty}frac{(-1)^n}{binom{2n}{n}(2n+1)}x^n
end{align*}
which you can show equals
begin{align*}
g(x) = 4 frac{sinh^{-1}(sqrt{x}/2)}{sqrt{x(x+4)}}
end{align*}
If we define $L = x frac{d}{dx}$ as the operator which first takes the derivative w.r.t $x$, then multiplies by $x$, we get that
begin{align*}
C = (L^0 g)(1), qquad B = (L^1 g)(1), qquad A = (L^2 g)(1)
end{align*}
or
begin{align*}
C = 4 frac{sinh^{-1}(1/2)}{sqrt{5}}, qquad B = frac{2}{5} - frac{12 sinh^{-1}(1/2)}{5sqrt{5}}, qquad A = frac{4}{125}(7sqrt{5}sinh^{-1}(1/2) - 10)
end{align*}






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Awesome answer! How did you come up with the generating functions and the corresponding closed form solution? Can you propose any resource for this?
    $endgroup$
    – MrYouMath
    2 hours ago






  • 1




    $begingroup$
    @MrYouMath: If you write $binom{2n}{n}=binom{2n}{2n-n}$, then the generating function $g(x)$ is the convolution of coefficients $a_n:=binom{2n}{n}^{-1}$ and $b_n:=(2n+1)^{-1}$, which is equivalent to the product of their individual generating functions.
    $endgroup$
    – Alex R.
    1 hour ago












  • $begingroup$
    Hi @MrYouMath, Alex's method is perhaps the best way to discover the generating function. Sometimes, you can even recognize the coefficients to be similar to existing Taylor expansions. In this case, the $a_n b_n$ collectively has similar form to the coefficients in the Taylor series expansion of $sinh^{-1}(x)$. From there, you can apply formal power series operations to try and and match the coefficients. The standard reference is generatingfunctionology by Wilf. Stanley's Enumerative Combinatorics also has an excellent presentation on generating functions / formal power series.
    $endgroup$
    – Tom Chen
    44 mins ago














6












6








6





$begingroup$

If we use Wolfram for these sums, with approximate values
begin{align*}
C approx 0.86082, qquad B approx -0.11649, qquad A approx -0.07897
end{align*}
From there, you can collect coefficients of $F_r$ to prove your identity. A more concrete way to derive exact expressions is through generating functions. Define
begin{align*}
g(x) = sum_{n=0}^{infty}frac{(-1)^n}{binom{2n}{n}(2n+1)}x^n
end{align*}
which you can show equals
begin{align*}
g(x) = 4 frac{sinh^{-1}(sqrt{x}/2)}{sqrt{x(x+4)}}
end{align*}
If we define $L = x frac{d}{dx}$ as the operator which first takes the derivative w.r.t $x$, then multiplies by $x$, we get that
begin{align*}
C = (L^0 g)(1), qquad B = (L^1 g)(1), qquad A = (L^2 g)(1)
end{align*}
or
begin{align*}
C = 4 frac{sinh^{-1}(1/2)}{sqrt{5}}, qquad B = frac{2}{5} - frac{12 sinh^{-1}(1/2)}{5sqrt{5}}, qquad A = frac{4}{125}(7sqrt{5}sinh^{-1}(1/2) - 10)
end{align*}






share|cite|improve this answer











$endgroup$



If we use Wolfram for these sums, with approximate values
begin{align*}
C approx 0.86082, qquad B approx -0.11649, qquad A approx -0.07897
end{align*}
From there, you can collect coefficients of $F_r$ to prove your identity. A more concrete way to derive exact expressions is through generating functions. Define
begin{align*}
g(x) = sum_{n=0}^{infty}frac{(-1)^n}{binom{2n}{n}(2n+1)}x^n
end{align*}
which you can show equals
begin{align*}
g(x) = 4 frac{sinh^{-1}(sqrt{x}/2)}{sqrt{x(x+4)}}
end{align*}
If we define $L = x frac{d}{dx}$ as the operator which first takes the derivative w.r.t $x$, then multiplies by $x$, we get that
begin{align*}
C = (L^0 g)(1), qquad B = (L^1 g)(1), qquad A = (L^2 g)(1)
end{align*}
or
begin{align*}
C = 4 frac{sinh^{-1}(1/2)}{sqrt{5}}, qquad B = frac{2}{5} - frac{12 sinh^{-1}(1/2)}{5sqrt{5}}, qquad A = frac{4}{125}(7sqrt{5}sinh^{-1}(1/2) - 10)
end{align*}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 3 hours ago









Tom ChenTom Chen

858412




858412








  • 1




    $begingroup$
    Awesome answer! How did you come up with the generating functions and the corresponding closed form solution? Can you propose any resource for this?
    $endgroup$
    – MrYouMath
    2 hours ago






  • 1




    $begingroup$
    @MrYouMath: If you write $binom{2n}{n}=binom{2n}{2n-n}$, then the generating function $g(x)$ is the convolution of coefficients $a_n:=binom{2n}{n}^{-1}$ and $b_n:=(2n+1)^{-1}$, which is equivalent to the product of their individual generating functions.
    $endgroup$
    – Alex R.
    1 hour ago












  • $begingroup$
    Hi @MrYouMath, Alex's method is perhaps the best way to discover the generating function. Sometimes, you can even recognize the coefficients to be similar to existing Taylor expansions. In this case, the $a_n b_n$ collectively has similar form to the coefficients in the Taylor series expansion of $sinh^{-1}(x)$. From there, you can apply formal power series operations to try and and match the coefficients. The standard reference is generatingfunctionology by Wilf. Stanley's Enumerative Combinatorics also has an excellent presentation on generating functions / formal power series.
    $endgroup$
    – Tom Chen
    44 mins ago














  • 1




    $begingroup$
    Awesome answer! How did you come up with the generating functions and the corresponding closed form solution? Can you propose any resource for this?
    $endgroup$
    – MrYouMath
    2 hours ago






  • 1




    $begingroup$
    @MrYouMath: If you write $binom{2n}{n}=binom{2n}{2n-n}$, then the generating function $g(x)$ is the convolution of coefficients $a_n:=binom{2n}{n}^{-1}$ and $b_n:=(2n+1)^{-1}$, which is equivalent to the product of their individual generating functions.
    $endgroup$
    – Alex R.
    1 hour ago












  • $begingroup$
    Hi @MrYouMath, Alex's method is perhaps the best way to discover the generating function. Sometimes, you can even recognize the coefficients to be similar to existing Taylor expansions. In this case, the $a_n b_n$ collectively has similar form to the coefficients in the Taylor series expansion of $sinh^{-1}(x)$. From there, you can apply formal power series operations to try and and match the coefficients. The standard reference is generatingfunctionology by Wilf. Stanley's Enumerative Combinatorics also has an excellent presentation on generating functions / formal power series.
    $endgroup$
    – Tom Chen
    44 mins ago








1




1




$begingroup$
Awesome answer! How did you come up with the generating functions and the corresponding closed form solution? Can you propose any resource for this?
$endgroup$
– MrYouMath
2 hours ago




$begingroup$
Awesome answer! How did you come up with the generating functions and the corresponding closed form solution? Can you propose any resource for this?
$endgroup$
– MrYouMath
2 hours ago




1




1




$begingroup$
@MrYouMath: If you write $binom{2n}{n}=binom{2n}{2n-n}$, then the generating function $g(x)$ is the convolution of coefficients $a_n:=binom{2n}{n}^{-1}$ and $b_n:=(2n+1)^{-1}$, which is equivalent to the product of their individual generating functions.
$endgroup$
– Alex R.
1 hour ago






$begingroup$
@MrYouMath: If you write $binom{2n}{n}=binom{2n}{2n-n}$, then the generating function $g(x)$ is the convolution of coefficients $a_n:=binom{2n}{n}^{-1}$ and $b_n:=(2n+1)^{-1}$, which is equivalent to the product of their individual generating functions.
$endgroup$
– Alex R.
1 hour ago














$begingroup$
Hi @MrYouMath, Alex's method is perhaps the best way to discover the generating function. Sometimes, you can even recognize the coefficients to be similar to existing Taylor expansions. In this case, the $a_n b_n$ collectively has similar form to the coefficients in the Taylor series expansion of $sinh^{-1}(x)$. From there, you can apply formal power series operations to try and and match the coefficients. The standard reference is generatingfunctionology by Wilf. Stanley's Enumerative Combinatorics also has an excellent presentation on generating functions / formal power series.
$endgroup$
– Tom Chen
44 mins ago




$begingroup$
Hi @MrYouMath, Alex's method is perhaps the best way to discover the generating function. Sometimes, you can even recognize the coefficients to be similar to existing Taylor expansions. In this case, the $a_n b_n$ collectively has similar form to the coefficients in the Taylor series expansion of $sinh^{-1}(x)$. From there, you can apply formal power series operations to try and and match the coefficients. The standard reference is generatingfunctionology by Wilf. Stanley's Enumerative Combinatorics also has an excellent presentation on generating functions / formal power series.
$endgroup$
– Tom Chen
44 mins ago










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Unua mondmilito

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Unua mondmilito Daŭro: 28-a de julio 1914 - 11-a de novembro 1918 Loko: Mondmilito (Eŭropo, Afriko kaj Proksima Oriento (mallongtempe en Ĉinio kaj Pacifiko) Rezulto: Venko de Interkonsento. La fino de la germana, la rusia kaj la aŭstra-hungara imperioj, ankaŭ la Osmanida imperio. La estiĝo de novaj landoj en la meza kaj la orienta Eŭropo. Casus belli : Murdo de sekvonto de trono Francisko Ferdinando (la 28-an de junio), la sekva aŭstria proklamo de milito al Serbio (la 28-an de julio), la rusa mobilizo kontraŭ Aŭstrio-Hungario (la 29-an de julio). Ankaŭ la naciismo, militismo kaj imperiismo. Flankoj Interkonsento : Cara Rusio Serba reĝlando Francio Brita imperio Itala reĝlando Usono Japanio Centraj potencoj : Germana imperio Aŭstrio-Hungario Osmanida imperio Bulgara imperio Komandantoj Georges Clemenceau Joseph Joffre Ferdinand Foch Robert Nivelle Herbert H. Asquith D. Lloyd George Douglas Haig John Jellicoe Vi...
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Schloss Hohenburg (Lenggries)

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Schloss Hohenburg Schloss Hohenburg ist ein Gemeindeteil der Gemeinde Lenggries im Isarwinkel. [1] Inhaltsverzeichnis 1 Baugeschichte 2 Bauherr 2.1 Spätere Eigentümer 3 Fotogalerie 4 Einzelnachweise 5 Literatur 6 Weblinks Baugeschichte | Bereits seit dem 11. Jahrhundert gab es bei Lenggries die mächtige Burg Hohenburg, von wo jahrhundertelang die Geschicke des Isarwinkels geleitet wurden. Seit 1566 war diese Burg im Besitz der Augsburger Adelsfamilie und Patrizier Herwarth . Während des Bayerischen Volksaufstandes von 1705 stand die stolze Veste im Zentrum dieser Erhebung, anschließend, im Spanischen Erbfolgekrieg wurde die Burg von österreichischen Soldaten besetzt. Am 21. Juli 1707 ging die Burg, unter bis heute ungeklärten Umständen, in Flammen auf und wurde in der Feuersbrunst zerstört (Genaueres zur Burg im Hauptartikel Hohenburg (Lenggries)). Bauherr | Graf Ferdinand Joseph von Herwarth , der zu dem Zeitpunkt Herr der Burg...
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