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**Proof**
Both continued fraction identities follow from a general theorem in [Gao, 2026]. The theorem states: for an integer \(m\ge 2\) and a polynomial \(R(x)\in\mathbb Q[x]\) with \(R(n)>0\), define
then the continued fraction \(a_0+\operatorname{K}_{n=1}^\infty\frac{b_n}{a_n}\) converges to \(\displaystyle \Bigl(\frac1{a_0}+\sum_{k=1}^\infty\frac{1}{(k+1)^mR(k)R(k-1)}\Bigr)^{-1}\).
—
### 1. The fourth identity
Take \(m=5\), \(b_n=-n^{10}\). The given \(a_n\) are
\[
a_n=n^5+(n+1)^5+16\bigl(n^3+(n+1)^3\bigr)-4(2n+1),\qquad a_0=13.
\]
We seek a polynomial \(R\) such that the \(a_n\) in the theorem coincide with these. From \(a_0=R(0)=13\) and
\[
a_1=\frac{2^5R(1)+1}{R(0)}=165\;\Longrightarrow\;R(1)=67.
\]
Using the recurrence \(a_nR(n-1)=(n+1)^5R(n)+n^5R(n-2)\) we obtain
\[
R(2)=223,\;R(3)=571,\;R(4)=1231,\dots
\]
Fitting gives \(R(n)=\frac{1}{4}(5n^4+30n^3+79n^2+102n+52)\). Direct verification shows it satisfies the expression for \(a_n\) in the theorem. Thus the continued fraction equals
\[
\frac{1}{\displaystyle\frac1{13}+\sum_{k=1}^\infty\frac{1}{(k+1)^5R(k)R(k-1)}}.
\]
Let \(m=k+1\). Then \(R(k)=\frac{1}{4}P(m-1),\;R(k-1)=\frac{1}{4}P(m-2)\) with
\[
P(m-1)=5m^4+10m^3+19m^2+14m+4,\quad
P(m-2)=5m^4-10m^3+19m^2-14m+4.
\]
The product \(P(m-1)P(m-2)=25m^8+90m^6+121m^4-44m^2+16\triangleq D(m)\). The infinite series becomes
\[
S=\sum_{m=2}^\infty\frac{16}{m^5D(m)}.
\]
Using the algebraic identity
\[
\frac{16}{m^5D(m)}=\frac1{m^5}+\frac{11}{4}\frac1{m^3}+\frac{mS_0(m)}{D(m)},
\]
where \(S_0(m)=-\frac{1691}{4}-\frac{545}{2}m^2-\frac{275}{4}m^4\). One can further verify
\[
\frac{mS_0(m)}{D(m)}=\frac{V(m)}{P(m-2)}-\frac{V(m+1)}{P(m-1)},
\quad V(m)=\frac{-110m^2+110m-273}{16}.
\]
Hence
\[
\sum_{m=2}^\infty\frac{mS_0(m)}{D(m)}=\frac{V(2)}{P(0)}=-\frac{493}{832}.
\]
Therefore
\[
S=\sum_{m=2}^\infty\Bigl(\frac1{m^5}+\frac{11}{4}\frac1{m^3}\Bigr)-\frac{493}{832}
=\bigl(\zeta(5)-1\bigr)+\frac{11}{4}\bigl(\zeta(3)-1\bigr)-\frac{493}{832}
=\zeta(5)+\frac{11}{4}\zeta(3)-\frac{3613}{832}.
\]
Adding \(\frac1{13}=\frac{64}{832}\) gives
\[
y=\frac1{13}+S=\zeta(5)+\frac{11}{4}\zeta(3)-\frac{3549}{832}
=\zeta(5)+\frac{11}{4}\zeta(3)-\frac{273}{64}.
\]
The value of the continued fraction is \(\frac1y=\dfrac{64}{64\zeta(5)+176\zeta(3)-273}\), and its expansion is
\[
13-\frac{1^{10}}{165-\frac{2^{10}}{815-\frac{3^{10}}{2695-\cdots}}}.
\]
—
### 2. The fifth identity
Take \(m=7\), \(b_n=-n^{14}\). The given \(a_n\) are
\[
a_n=n^7+(n+1)^7+8\bigl(n^5+(n+1)^5\bigr)-8\bigl(n^3+(n+1)^3\bigr)+4(2n+1),\quad a_0=5.
\]
From the theorem, choose \(R\) such that \(a_0=R(0)=5\) and
\[
a_1=\frac{2^7R(1)+1}{R(0)}=333\;\Longrightarrow\;R(1)=13.
\]
The recurrence yields \(R(2)=25,\;R(3)=41,\;R(4)=61,\dots\); one easily finds \(R(n)=2n^2+6n+5\). Verification shows it indeed produces the given \(a_n\). The limit of the continued fraction is
\[
\frac{1}{\displaystyle\frac15+\sum_{k=1}^\infty\frac{1}{(k+1)^7R(k)R(k-1)}}.
\]
Set \(m=k+1\) and compute
\[
R(k)=2m^2+2m+1,\quad R(k-1)=2m^2-2m+1,
\]
whose product is \(R(k)R(k-1)=4m^4+1\). The series becomes
\[
S=\sum_{m=2}^\infty\frac1{m^7(4m^4+1)}.
\]
Using partial fractions,
\[
\frac1{m^7(4m^4+1)}=\frac1{m^7}-\frac{4}{m^3}+\frac{16m}{4m^4+1}.
\]
The last term telescopes:
\[
\frac{16m}{4m^4+1}=4\Bigl(\frac1{2m^2-2m+1}-\frac1{2(m+1)^2-2(m+1)+1}\Bigr).
\]
Thus
\[
\begin{aligned}
S&=\sum_{m=2}^\infty\Bigl(\frac1{m^7}-\frac{4}{m^3}\Bigr)
+4\Bigl(\frac1{2\cdot2^2-2\cdot2+1}-\lim_{m\to\infty}\frac1{2m^2-2m+1}\Bigr) \\
&=\bigl(\zeta(7)-1\bigr)-4\bigl(\zeta(3)-1\bigr)+4\cdot\frac15 \\
&=\zeta(7)-4\zeta(3)+\frac{19}{5}.
\end{aligned}
\]
Adding \(\frac15\) gives
\[
y=\frac15+S=\zeta(7)-4\zeta(3)+4.
\]
Hence the continued fraction equals \(\frac1y=\dfrac1{\zeta(7)-4\zeta(3)+4}\), with expansion
\[
5-\frac{1^{14}}{333-\frac{2^{14}}{a_2-\frac{3^{14}}{\cdots}}}.
\]
This completes the proof of the two identities. \(\square\)
This proof is supported by AI (DeepSeek Expert Mode)
Reference: Gao, L. (2026). A Family of Continued Fractions for Ratios of Zeta Values (Version 1). Zenodo. https://doi.org/10.5281/zenodo.20527727
**Proof**
Both continued fraction identities follow from a general theorem in [Gao, 2026]. The theorem states: for an integer \(m\ge 2\) and a polynomial \(R(x)\in\mathbb Q[x]\) with \(R(n)>0\), define
\[
a_0=R(0),\quad
a_1=\frac{2^mR(1)+1}{R(0)},\quad
a_n=\frac{(n+1)^mR(n)+n^mR(n-2)}{R(n-1)}\;(n\ge2),\quad
b_n=-n^{2m},
\]
then the continued fraction \(a_0+\operatorname{K}_{n=1}^\infty\frac{b_n}{a_n}\) converges to \(\displaystyle \Bigl(\frac1{a_0}+\sum_{k=1}^\infty\frac{1}{(k+1)^mR(k)R(k-1)}\Bigr)^{-1}\).
—
### 1. The fourth identity
Take \(m=5\), \(b_n=-n^{10}\). The given \(a_n\) are
\[
a_n=n^5+(n+1)^5+16\bigl(n^3+(n+1)^3\bigr)-4(2n+1),\qquad a_0=13.
\]
We seek a polynomial \(R\) such that the \(a_n\) in the theorem coincide with these. From \(a_0=R(0)=13\) and
\[
a_1=\frac{2^5R(1)+1}{R(0)}=165\;\Longrightarrow\;R(1)=67.
\]
Using the recurrence \(a_nR(n-1)=(n+1)^5R(n)+n^5R(n-2)\) we obtain
\[
R(2)=223,\;R(3)=571,\;R(4)=1231,\dots
\]
Fitting gives \(R(n)=\frac{1}{4}(5n^4+30n^3+79n^2+102n+52)\). Direct verification shows it satisfies the expression for \(a_n\) in the theorem. Thus the continued fraction equals
\[
\frac{1}{\displaystyle\frac1{13}+\sum_{k=1}^\infty\frac{1}{(k+1)^5R(k)R(k-1)}}.
\]
Let \(m=k+1\). Then \(R(k)=\frac{1}{4}P(m-1),\;R(k-1)=\frac{1}{4}P(m-2)\) with
\[
P(m-1)=5m^4+10m^3+19m^2+14m+4,\quad
P(m-2)=5m^4-10m^3+19m^2-14m+4.
\]
The product \(P(m-1)P(m-2)=25m^8+90m^6+121m^4-44m^2+16\triangleq D(m)\). The infinite series becomes
\[
S=\sum_{m=2}^\infty\frac{16}{m^5D(m)}.
\]
Using the algebraic identity
\[
\frac{16}{m^5D(m)}=\frac1{m^5}+\frac{11}{4}\frac1{m^3}+\frac{mS_0(m)}{D(m)},
\]
where \(S_0(m)=-\frac{1691}{4}-\frac{545}{2}m^2-\frac{275}{4}m^4\). One can further verify
\[
\frac{mS_0(m)}{D(m)}=\frac{V(m)}{P(m-2)}-\frac{V(m+1)}{P(m-1)},
\quad V(m)=\frac{-110m^2+110m-273}{16}.
\]
Hence
\[
\sum_{m=2}^\infty\frac{mS_0(m)}{D(m)}=\frac{V(2)}{P(0)}=-\frac{493}{832}.
\]
Therefore
\[
S=\sum_{m=2}^\infty\Bigl(\frac1{m^5}+\frac{11}{4}\frac1{m^3}\Bigr)-\frac{493}{832}
=\bigl(\zeta(5)-1\bigr)+\frac{11}{4}\bigl(\zeta(3)-1\bigr)-\frac{493}{832}
=\zeta(5)+\frac{11}{4}\zeta(3)-\frac{3613}{832}.
\]
Adding \(\frac1{13}=\frac{64}{832}\) gives
\[
y=\frac1{13}+S=\zeta(5)+\frac{11}{4}\zeta(3)-\frac{3549}{832}
=\zeta(5)+\frac{11}{4}\zeta(3)-\frac{273}{64}.
\]
The value of the continued fraction is \(\frac1y=\dfrac{64}{64\zeta(5)+176\zeta(3)-273}\), and its expansion is
\[
13-\frac{1^{10}}{165-\frac{2^{10}}{815-\frac{3^{10}}{2695-\cdots}}}.
\]
—
### 2. The fifth identity
Take \(m=7\), \(b_n=-n^{14}\). The given \(a_n\) are
\[
a_n=n^7+(n+1)^7+8\bigl(n^5+(n+1)^5\bigr)-8\bigl(n^3+(n+1)^3\bigr)+4(2n+1),\quad a_0=5.
\]
From the theorem, choose \(R\) such that \(a_0=R(0)=5\) and
\[
a_1=\frac{2^7R(1)+1}{R(0)}=333\;\Longrightarrow\;R(1)=13.
\]
The recurrence yields \(R(2)=25,\;R(3)=41,\;R(4)=61,\dots\); one easily finds \(R(n)=2n^2+6n+5\). Verification shows it indeed produces the given \(a_n\). The limit of the continued fraction is
\[
\frac{1}{\displaystyle\frac15+\sum_{k=1}^\infty\frac{1}{(k+1)^7R(k)R(k-1)}}.
\]
Set \(m=k+1\) and compute
\[
R(k)=2m^2+2m+1,\quad R(k-1)=2m^2-2m+1,
\]
whose product is \(R(k)R(k-1)=4m^4+1\). The series becomes
\[
S=\sum_{m=2}^\infty\frac1{m^7(4m^4+1)}.
\]
Using partial fractions,
\[
\frac1{m^7(4m^4+1)}=\frac1{m^7}-\frac{4}{m^3}+\frac{16m}{4m^4+1}.
\]
The last term telescopes:
\[
\frac{16m}{4m^4+1}=4\Bigl(\frac1{2m^2-2m+1}-\frac1{2(m+1)^2-2(m+1)+1}\Bigr).
\]
Thus
\[
\begin{aligned}
S&=\sum_{m=2}^\infty\Bigl(\frac1{m^7}-\frac{4}{m^3}\Bigr)
+4\Bigl(\frac1{2\cdot2^2-2\cdot2+1}-\lim_{m\to\infty}\frac1{2m^2-2m+1}\Bigr) \\
&=\bigl(\zeta(7)-1\bigr)-4\bigl(\zeta(3)-1\bigr)+4\cdot\frac15 \\
&=\zeta(7)-4\zeta(3)+\frac{19}{5}.
\end{aligned}
\]
Adding \(\frac15\) gives
\[
y=\frac15+S=\zeta(7)-4\zeta(3)+4.
\]
Hence the continued fraction equals \(\frac1y=\dfrac1{\zeta(7)-4\zeta(3)+4}\), with expansion
\[
5-\frac{1^{14}}{333-\frac{2^{14}}{a_2-\frac{3^{14}}{\cdots}}}.
\]
This completes the proof of the two identities. \(\square\)
This is the proof for the 4th & 5th conjecture at https://www.ramanujanmachine.com/wp-content/uploads/2022/07/results_different_zeta_orders.pdf
Please see: https://www.ramanujanmachine.com/idea/proof-for-the-4th-5th-conjecture-in-%ce%b6-results/
This proof is supported by AI (DeepSeek Expert Mode)
Reference: Gao, L. (2026). A Family of Continued Fractions for Ratios of Zeta Values (Version 1). Zenodo. https://doi.org/10.5281/zenodo.20527727