Proof of Generalised ratio test












1














Is the following argument correct?



Proposition. Suppose ${x_n}$ is a sequence and suppose for some $xinmathbf{R}$, the limit
$$L:=lim_{ntoinfty}frac{|x_{n+1}-x|}{|x_n-x|}$$
exists and $L<1$. Show that ${x_n}$ converges to $x$.




Proof. We choose an $r$ such that $L<r<1$, then in particular for $epsilon = r-L$, there exists an $Minmathbf{N}$, such that
$left|frac{|x_{n+1}-x|}{|x_n-x|}-Lright|<r-L,forall nge M$ and by
extension $frac{|x_{n+1}-x|}{|x_n-x|}-L<r-L,forall nge M$. We may
therefore surmise that $$frac{|x_{n+1}-x|}{|x_n-x|}<r,forall nge
M$$ Now given an arbitrary $n>M$, the above proposition implies the
following begin{align*} |x_n-x| &= |x_M-x|frac{|x_{M+1}-x|}{|x_{M}-x|}frac{|x_{M+2}-x|}{|x_{M+1}-x|}cdotsfrac{|x_n-x|}{|x_{n-1}-x|}\ &<|x_M-x|rrcdots r = |x_M-x|r^{n-M} = (|x_M-x|r^{-M})r^{n}.
end{align*} consequently the sequence ${x_{n+M}}_{n=1}^{infty}$ is
such that we have $$|x_{n+M}-x|< (|x_M-x|r^{-M})r^{n},forall nin N$$
Now since $r<1$, $lim_{ntoinfty}r^n = 0$ and by extension
$lim_{ntoinfty}(|x_M-x|r^{-M})r^{n}= 0$. Finally appealing to
proposition $textbf{2.2.4}$ implies that ${x_{n+M}}_{n=1}^{infty}$
and by extension ${x_n}$ converge to $x$.



$blacksquare$




Note: Proposition $textbf{2.2.4}$ is the result that if $xinmathbf{R}$
and we have a sequence ${a_n}$ such that $lim_{ntoinfty}a_n = 0$,and we have a sequence $|x_n-x|<a_n,forall ninmathbf{N}$, then $lim_{ntoinfty}x_n = x$.






real-analysis sequences-and-series proof-verification







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  • Yes, the proof is correct. Good job!
    – Saad
    Oct 17 '18 at 2:40
















1














Is the following argument correct?



Proposition. Suppose ${x_n}$ is a sequence and suppose for some $xinmathbf{R}$, the limit
$$L:=lim_{ntoinfty}frac{|x_{n+1}-x|}{|x_n-x|}$$
exists and $L<1$. Show that ${x_n}$ converges to $x$.




Proof. We choose an $r$ such that $L<r<1$, then in particular for $epsilon = r-L$, there exists an $Minmathbf{N}$, such that
$left|frac{|x_{n+1}-x|}{|x_n-x|}-Lright|<r-L,forall nge M$ and by
extension $frac{|x_{n+1}-x|}{|x_n-x|}-L<r-L,forall nge M$. We may
therefore surmise that $$frac{|x_{n+1}-x|}{|x_n-x|}<r,forall nge
M$$ Now given an arbitrary $n>M$, the above proposition implies the
following begin{align*} |x_n-x| &= |x_M-x|frac{|x_{M+1}-x|}{|x_{M}-x|}frac{|x_{M+2}-x|}{|x_{M+1}-x|}cdotsfrac{|x_n-x|}{|x_{n-1}-x|}\ &<|x_M-x|rrcdots r = |x_M-x|r^{n-M} = (|x_M-x|r^{-M})r^{n}.
end{align*} consequently the sequence ${x_{n+M}}_{n=1}^{infty}$ is
such that we have $$|x_{n+M}-x|< (|x_M-x|r^{-M})r^{n},forall nin N$$
Now since $r<1$, $lim_{ntoinfty}r^n = 0$ and by extension
$lim_{ntoinfty}(|x_M-x|r^{-M})r^{n}= 0$. Finally appealing to
proposition $textbf{2.2.4}$ implies that ${x_{n+M}}_{n=1}^{infty}$
and by extension ${x_n}$ converge to $x$.



$blacksquare$




Note: Proposition $textbf{2.2.4}$ is the result that if $xinmathbf{R}$
and we have a sequence ${a_n}$ such that $lim_{ntoinfty}a_n = 0$,and we have a sequence $|x_n-x|<a_n,forall ninmathbf{N}$, then $lim_{ntoinfty}x_n = x$.






real-analysis sequences-and-series proof-verification







share|cite|improve this question






















  • Yes, the proof is correct. Good job!
    – Saad
    Oct 17 '18 at 2:40














1












1








1







Is the following argument correct?



Proposition. Suppose ${x_n}$ is a sequence and suppose for some $xinmathbf{R}$, the limit
$$L:=lim_{ntoinfty}frac{|x_{n+1}-x|}{|x_n-x|}$$
exists and $L<1$. Show that ${x_n}$ converges to $x$.




Proof. We choose an $r$ such that $L<r<1$, then in particular for $epsilon = r-L$, there exists an $Minmathbf{N}$, such that
$left|frac{|x_{n+1}-x|}{|x_n-x|}-Lright|<r-L,forall nge M$ and by
extension $frac{|x_{n+1}-x|}{|x_n-x|}-L<r-L,forall nge M$. We may
therefore surmise that $$frac{|x_{n+1}-x|}{|x_n-x|}<r,forall nge
M$$ Now given an arbitrary $n>M$, the above proposition implies the
following begin{align*} |x_n-x| &= |x_M-x|frac{|x_{M+1}-x|}{|x_{M}-x|}frac{|x_{M+2}-x|}{|x_{M+1}-x|}cdotsfrac{|x_n-x|}{|x_{n-1}-x|}\ &<|x_M-x|rrcdots r = |x_M-x|r^{n-M} = (|x_M-x|r^{-M})r^{n}.
end{align*} consequently the sequence ${x_{n+M}}_{n=1}^{infty}$ is
such that we have $$|x_{n+M}-x|< (|x_M-x|r^{-M})r^{n},forall nin N$$
Now since $r<1$, $lim_{ntoinfty}r^n = 0$ and by extension
$lim_{ntoinfty}(|x_M-x|r^{-M})r^{n}= 0$. Finally appealing to
proposition $textbf{2.2.4}$ implies that ${x_{n+M}}_{n=1}^{infty}$
and by extension ${x_n}$ converge to $x$.



$blacksquare$




Note: Proposition $textbf{2.2.4}$ is the result that if $xinmathbf{R}$
and we have a sequence ${a_n}$ such that $lim_{ntoinfty}a_n = 0$,and we have a sequence $|x_n-x|<a_n,forall ninmathbf{N}$, then $lim_{ntoinfty}x_n = x$.






real-analysis sequences-and-series proof-verification







share|cite|improve this question













Is the following argument correct?



Proposition. Suppose ${x_n}$ is a sequence and suppose for some $xinmathbf{R}$, the limit
$$L:=lim_{ntoinfty}frac{|x_{n+1}-x|}{|x_n-x|}$$
exists and $L<1$. Show that ${x_n}$ converges to $x$.




Proof. We choose an $r$ such that $L<r<1$, then in particular for $epsilon = r-L$, there exists an $Minmathbf{N}$, such that
$left|frac{|x_{n+1}-x|}{|x_n-x|}-Lright|<r-L,forall nge M$ and by
extension $frac{|x_{n+1}-x|}{|x_n-x|}-L<r-L,forall nge M$. We may
therefore surmise that $$frac{|x_{n+1}-x|}{|x_n-x|}<r,forall nge
M$$ Now given an arbitrary $n>M$, the above proposition implies the
following begin{align*} |x_n-x| &= |x_M-x|frac{|x_{M+1}-x|}{|x_{M}-x|}frac{|x_{M+2}-x|}{|x_{M+1}-x|}cdotsfrac{|x_n-x|}{|x_{n-1}-x|}\ &<|x_M-x|rrcdots r = |x_M-x|r^{n-M} = (|x_M-x|r^{-M})r^{n}.
end{align*} consequently the sequence ${x_{n+M}}_{n=1}^{infty}$ is
such that we have $$|x_{n+M}-x|< (|x_M-x|r^{-M})r^{n},forall nin N$$
Now since $r<1$, $lim_{ntoinfty}r^n = 0$ and by extension
$lim_{ntoinfty}(|x_M-x|r^{-M})r^{n}= 0$. Finally appealing to
proposition $textbf{2.2.4}$ implies that ${x_{n+M}}_{n=1}^{infty}$
and by extension ${x_n}$ converge to $x$.



$blacksquare$




Note: Proposition $textbf{2.2.4}$ is the result that if $xinmathbf{R}$
and we have a sequence ${a_n}$ such that $lim_{ntoinfty}a_n = 0$,and we have a sequence $|x_n-x|<a_n,forall ninmathbf{N}$, then $lim_{ntoinfty}x_n = x$.






real-analysis sequences-and-series proof-verification




real-analysis sequences-and-series proof-verification






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asked Oct 17 '18 at 0:47









Atif FarooqAtif Farooq

3,1422825




3,1422825












  • Yes, the proof is correct. Good job!
    – Saad
    Oct 17 '18 at 2:40


















  • Yes, the proof is correct. Good job!
    – Saad
    Oct 17 '18 at 2:40
















Yes, the proof is correct. Good job!
– Saad
Oct 17 '18 at 2:40




Yes, the proof is correct. Good job!
– Saad
Oct 17 '18 at 2:40










1 Answer
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Yes. Your proof is right, but you also can use another simpler proof. Define $y_n=|x_n-x|$. Then according to Ratio test for $y_n$ we know that $sum y_n$ exists and is bounded since $$0<lim_{nto infty}{y_{n+1}over y_n}=L<1$$because $sum y_n$ exists and is bounded it must be necessarily that $y_nto 0$ or equivalently $x_nto x$






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    Yes. Your proof is right, but you also can use another simpler proof. Define $y_n=|x_n-x|$. Then according to Ratio test for $y_n$ we know that $sum y_n$ exists and is bounded since $$0<lim_{nto infty}{y_{n+1}over y_n}=L<1$$because $sum y_n$ exists and is bounded it must be necessarily that $y_nto 0$ or equivalently $x_nto x$






    share|cite|improve this answer


























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      Yes. Your proof is right, but you also can use another simpler proof. Define $y_n=|x_n-x|$. Then according to Ratio test for $y_n$ we know that $sum y_n$ exists and is bounded since $$0<lim_{nto infty}{y_{n+1}over y_n}=L<1$$because $sum y_n$ exists and is bounded it must be necessarily that $y_nto 0$ or equivalently $x_nto x$






      share|cite|improve this answer
























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        Yes. Your proof is right, but you also can use another simpler proof. Define $y_n=|x_n-x|$. Then according to Ratio test for $y_n$ we know that $sum y_n$ exists and is bounded since $$0<lim_{nto infty}{y_{n+1}over y_n}=L<1$$because $sum y_n$ exists and is bounded it must be necessarily that $y_nto 0$ or equivalently $x_nto x$






        share|cite|improve this answer












        Yes. Your proof is right, but you also can use another simpler proof. Define $y_n=|x_n-x|$. Then according to Ratio test for $y_n$ we know that $sum y_n$ exists and is bounded since $$0<lim_{nto infty}{y_{n+1}over y_n}=L<1$$because $sum y_n$ exists and is bounded it must be necessarily that $y_nto 0$ or equivalently $x_nto x$







        share|cite|improve this answer












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        share|cite|improve this answer










        answered Nov 30 '18 at 19:53









        Mostafa AyazMostafa Ayaz

        14.5k3937




        14.5k3937






























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