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Iosif Pinelis
Iosif Pinelis
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Consider the substitutions \begin{equation*} x_n=n+1/2+y_n/n,\quad y_n=u_n+5/8. \end{equation*} Then $u_1=-9/8$ and \begin{equation*} u_{n+1}=f_n(u_n) \end{equation*} for $n\ge1$, where \begin{equation*} f_n(u):=\frac{-64 n^4 u-8 n^3 (4 u-13)+n^2 (56 u+115)+n (96 u+76)+4 (8 u+5)}{8 n^2 \left(8 n^2+4 n+8 u+5\right)}. \end{equation*} Define $c_n(u)$ by the identity \begin{equation*} f_n(u)=-u+\frac{13}{8n}+\frac{c_n(u)}{n^2}. \end{equation*}\begin{equation*} f_n(u)=-u+\frac{13}{8n}+\frac{c_n(u)}{n^2}, \end{equation*} so that \begin{equation*} c_n(u)=\frac{n^2 \left(64 u^2+96 u+63\right)+n (11-8 u)+4 (8 u+5)}{8 \left(8 n^2+4 n+8 u+5\right)}. \end{equation*} Then for $n\ge1$ \begin{equation*} u_{n+1}+u_n=\frac{13}{8n}+\frac{c_n(u)}{n^2} \tag{1} \end{equation*} and for $n\ge2$ \begin{equation*} u_{n+1}=f_n(f_{n-1}(u_{n-1}))=u_{n-1}-\frac{13}{8n(n-1)}+\frac{c_n(u_n)}{n^2} -\frac{c_{n-1}(u_{n-1})}{(n-1)^2}. \tag{2} \end{equation*}

Note that \begin{equation*} u_{101}=-0.54\ldots,\quad u_{102}=0.56\ldots, \tag{3} \end{equation*} and \begin{equation*} 0\le c_n(u)\le3 \end{equation*} if $n\ge10$ and $u\in[-6/10,8/10]$. Therefore and because for natural $m\ge102$ we have \begin{equation*} \sum_{n=m}^\infty\Big(\frac{13}{8n(n-1)}+\frac3{(n-1)^2}\Big)<\frac5{m-2}\le0.05, \end{equation*} it follows from (2) and (3) by induction that for all $n\ge101$ we have $u_n\in[-6/10,8/10]$ and hence $0\le c_n(u_n)\le3$. So, again by (2), the sequences $(u_{2m})$ and $(u_{2m+1})$ are Cauchy-convergent and hence convergent. Moreover, by (1), $u_{n+1}+u_n\to0$.

Thus, indeed \begin{equation*} y_{n+1}+y_n\to5/4, \end{equation*} and the sequences $(y_{2m})$ and $(y_{2m+1})$ are convergent. (The limits of these two sequences can in principle be found numerically with any degree of accuracy -- controlled by (2), say.)

Consider the substitutions \begin{equation*} x_n=n+1/2+y_n/n,\quad y_n=u_n+5/8. \end{equation*} Then $u_1=-9/8$ and \begin{equation*} u_{n+1}=f_n(u_n) \end{equation*} for $n\ge1$, where \begin{equation*} f_n(u):=\frac{-64 n^4 u-8 n^3 (4 u-13)+n^2 (56 u+115)+n (96 u+76)+4 (8 u+5)}{8 n^2 \left(8 n^2+4 n+8 u+5\right)}. \end{equation*} Define $c_n(u)$ by the identity \begin{equation*} f_n(u)=-u+\frac{13}{8n}+\frac{c_n(u)}{n^2}. \end{equation*} Then for $n\ge1$ \begin{equation*} u_{n+1}+u_n=\frac{13}{8n}+\frac{c_n(u)}{n^2} \tag{1} \end{equation*} and for $n\ge2$ \begin{equation*} u_{n+1}=f_n(f_{n-1}(u_{n-1}))=u_{n-1}-\frac{13}{8n(n-1)}+\frac{c_n(u_n)}{n^2} -\frac{c_{n-1}(u_{n-1})}{(n-1)^2}. \tag{2} \end{equation*}

Note that \begin{equation*} u_{101}=-0.54\ldots,\quad u_{102}=0.56\ldots, \tag{3} \end{equation*} and \begin{equation*} 0\le c_n(u)\le3 \end{equation*} if $n\ge10$ and $u\in[-6/10,8/10]$. Therefore and because for natural $m\ge102$ we have \begin{equation*} \sum_{n=m}^\infty\Big(\frac{13}{8n(n-1)}+\frac3{(n-1)^2}\Big)<\frac5{m-2}\le0.05, \end{equation*} it follows from (2) and (3) by induction that for all $n\ge101$ we have $u_n\in[-6/10,8/10]$ and hence $0\le c_n(u_n)\le3$. So, again by (2), the sequences $(u_{2m})$ and $(u_{2m+1})$ are Cauchy-convergent and hence convergent. Moreover, by (1), $u_{n+1}+u_n\to0$.

Thus, indeed \begin{equation*} y_{n+1}+y_n\to5/4, \end{equation*} and the sequences $(y_{2m})$ and $(y_{2m+1})$ are convergent. (The limits of these two sequences can in principle be found numerically with any degree of accuracy.)

Consider the substitutions \begin{equation*} x_n=n+1/2+y_n/n,\quad y_n=u_n+5/8. \end{equation*} Then $u_1=-9/8$ and \begin{equation*} u_{n+1}=f_n(u_n) \end{equation*} for $n\ge1$, where \begin{equation*} f_n(u):=\frac{-64 n^4 u-8 n^3 (4 u-13)+n^2 (56 u+115)+n (96 u+76)+4 (8 u+5)}{8 n^2 \left(8 n^2+4 n+8 u+5\right)}. \end{equation*} Define $c_n(u)$ by the identity \begin{equation*} f_n(u)=-u+\frac{13}{8n}+\frac{c_n(u)}{n^2}, \end{equation*} so that \begin{equation*} c_n(u)=\frac{n^2 \left(64 u^2+96 u+63\right)+n (11-8 u)+4 (8 u+5)}{8 \left(8 n^2+4 n+8 u+5\right)}. \end{equation*} Then for $n\ge1$ \begin{equation*} u_{n+1}+u_n=\frac{13}{8n}+\frac{c_n(u)}{n^2} \tag{1} \end{equation*} and for $n\ge2$ \begin{equation*} u_{n+1}=f_n(f_{n-1}(u_{n-1}))=u_{n-1}-\frac{13}{8n(n-1)}+\frac{c_n(u_n)}{n^2} -\frac{c_{n-1}(u_{n-1})}{(n-1)^2}. \tag{2} \end{equation*}

Note that \begin{equation*} u_{101}=-0.54\ldots,\quad u_{102}=0.56\ldots, \tag{3} \end{equation*} and \begin{equation*} 0\le c_n(u)\le3 \end{equation*} if $n\ge10$ and $u\in[-6/10,8/10]$. Therefore and because for natural $m\ge102$ we have \begin{equation*} \sum_{n=m}^\infty\Big(\frac{13}{8n(n-1)}+\frac3{(n-1)^2}\Big)<\frac5{m-2}\le0.05, \end{equation*} it follows from (2) and (3) by induction that for all $n\ge101$ we have $u_n\in[-6/10,8/10]$ and hence $0\le c_n(u_n)\le3$. So, again by (2), the sequences $(u_{2m})$ and $(u_{2m+1})$ are Cauchy-convergent and hence convergent. Moreover, by (1), $u_{n+1}+u_n\to0$.

Thus, indeed \begin{equation*} y_{n+1}+y_n\to5/4, \end{equation*} and the sequences $(y_{2m})$ and $(y_{2m+1})$ are convergent. (The limits of these two sequences can in principle be found numerically with any degree of accuracy -- controlled by (2), say.)

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Iosif Pinelis
Iosif Pinelis
  • 127.9k
  • 8
  • 107
  • 229

Consider the substitutions \begin{equation*} x_n=n+1/2+y_n/n,\quad y_n=u_n+5/8. \end{equation*} Then $u_1=-9/8$ and \begin{equation*} u_{n+1}=f_n(u_n) \end{equation*} for $n\ge1$, where \begin{equation*} f_n(u):=\frac{-64 n^4 u-8 n^3 (4 u-13)+n^2 (56 u+115)+n (96 u+76)+4 (8 u+5)}{8 n^2 \left(8 n^2+4 n+8 u+5\right)}. \end{equation*} Define $c_n(u)$ by the identity \begin{equation*} f_n(u)=-u+\frac{13}{8n}+\frac{c_n(u)}{n^2}. \end{equation*} Then for $n\ge1$ \begin{equation*} u_{n+1}+u_n=\frac{13}{8n}+\frac{c_n(u)}{n^2} \tag{1} \end{equation*} and for $n\ge2$ \begin{equation*} u_{n+1}=f_n(f_{n-1}(u_{n-1}))=u_{n-1}-\frac{13}{8n(n-1)}+\frac{c_n(u_n)}{n^2} -\frac{c_{n-1}(u_{n-1})}{(n-1)^2}. \tag{2} \end{equation*}

Note that \begin{equation*} u_{101}=-0.54\ldots,\quad u_{102}=0.56\ldots, \tag{3} \end{equation*} and \begin{equation*} 0\le c_n(u)\le3 \end{equation*} if $n\ge10$ and $u\in[-6/10,8/10]$. Therefore and because for natural $m\ge102$ we have \begin{equation*} \sum_{n=m}^\infty\Big(\frac{13}{8n(n-1)}+\frac3{(n-1)^2}\Big)<\frac5{m-2}\le0.05, \end{equation*} it follows from (2) and (3) by induction that for all $n\ge101$ we have $u_n\in[-6/10,8/10]$ and hence $0\le c_n(u_n)\le3$. So, again by (2), the sequences $(u_{2m})$ and $(u_{2m+1})$ are Cauchy-convergent and hence convergent. Moreover, by (1), $u_{n+1}+u_n\to0$.

Thus, indeed \begin{equation*} y_{n+1}+y_n\to5/4, \end{equation*} and the sequences $(y_{2m})$ and $(y_{2m+1})$ are convergent. (The limits of these two sequences can in principle be found numerically with any degree of accuracy.)