Order of convergence limit Let $(x_n)$ be a real valued sequence converging to $\overline{x}$, with order of convergence $k \in \mathbb{N}$. This means the limit $$ \lim_{n \to \infty}\frac{|x_{n+1} - \overline{x}|}{{|x_{n} - \overline{x}|}^k}$$
exists. What I have to prove is that this limit is also equal to the limit $$ \lim_{n \to \infty}\frac{|x_{n+1} - x_n|}{{|x_{n} - x_{n-1}|}^k}$$
that is, to prove $$ \lim_{n \to \infty}\frac{|x_{n+1} - \overline{x}|}{{|x_{n} - \overline{x}|}^k} = \lim_{n \to \infty}\frac{|x_{n+1} - x_n|}{{|x_{n} - x_{n-1}|}^k} $$
I have been trying algebraic manipulations for hours now, but to no avail. At first, I though this problem was going to use the fact that Cauchy sequences converge in $\mathbb{R}$, but this didn’t get me anywhere, so I then decided to try using algebra and absolute value inequalities (like the triangle inequality), but I couldn’t get the correct denominator to appear.
So I set $k=1$ to see if it would make it easier, but the problem wasn’t in the value of $k$ itself, but rather in the product that would show up when trying to go from one side to the other.
I have tried going from the left hand side to the right one, and vice-versa, and neither of these approaches got me anywhere...
Could anyone help me? What am I missing?
Thanks in advance!
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If\ \lim _{n\rightarrow \infty } x_{n} =L\ \ for\ some\ L\in R,\ then\ we\ let\ g( n) =x_{n} -L\ hence\ \lim _{n\rightarrow \infty } \ g( n) =0\\
\\
\\
If\ \lim _{n\rightarrow \infty } x_{n} =L\ \ for\ some\ L\in R,\\
then\ we\ can\ conclude\ that\ x_{n} -L=g( n) \ for\ some\ g\ where\ \lim _{n\rightarrow \infty } g( n) =0\ \\
the\ first\ limit,\lim _{n\rightarrow \infty }\frac{|x_{n+1} -L|}{|x_{n} -L|^{k}} =A\ for\ some\ A\in R\\
then\ becomes\lim _{n\rightarrow \infty }\frac{|g( n+1) |}{|g( n) |^{k}} =A\ \\
we\ conclude\ that\ \frac{|g( n+1) |}{|g( n) |^{k}} -A=f( n) \ for\ some\ f\ where\ \lim _{n\rightarrow \infty } f( n) =0\ \\
hence\ g( n+1) =[ g( n)]^{k} *[ A+f( n)] *( -1)^{S( n)} \ where\ S\ is\ some\ function\ N\rightarrow N\\
We\ need\ the\ ( -1)^{S( n)} \ function\ because\ we\ dont\ know\ how\ the\ signs\ of\ these\ two\\
sides\ relate\ to\ one\ another\ but\ we\ know\ they\ are\ equal\ in\ magintude.\\
\\
So\ now\ to\ look\ at\ the\ second\ limit:\\
LIMIT=\lim _{n\rightarrow \infty }\frac{|x_{n+1} -x_{n} |}{|x_{n} -x_{n-1} |^{k}} =\lim _{n\rightarrow \infty }\frac{|g( n+1) -g( n) |}{|g( n) -g( n-1) |^{k}}\\
=\lim _{n\rightarrow \infty }\frac{|[ g( n)]^{k} *[ A+f( n)] *( -1)^{S( n)} -g( n) |}{|[ g( n-1)]^{k} *[ A+f( n-1)] *( -1)^{S( n-1)} -g( n-1) |^{k}}\\
\\
factoring\ out\ we\ get:\\
=\lim _{n\rightarrow \infty }\frac{|[ g( n)] |*|\ [ g( n)]^{k-1} *[ A+f( n)] *( -1)^{S( n)} -1\ |}{|[ g( n-1)]^{k} |*|\ [ g( n-1)]^{k-1} *[ A+f( n-1)] *( -1)^{S( n-1)} -1|^{k}}\\
\\
but\ we\ know\ \lim _{n\rightarrow \infty }\frac{|g( n+1) |}{|g( n) |^{k}} =A\ \ hence\ \lim _{n\rightarrow \infty }\frac{|g( n) |}{|g( n-1) |^{k}} =A\ and\ so\\
\\
LIMIT=A*\lim _{n\rightarrow \infty }\frac{|\ [ g( n)]^{k-1} *[ A+f( n)] *( -1)^{S( n)} -1\ |}{|\ [ g( n-1)]^{k-1} *[ A+f( n-1)] *( -1)^{S( n-1)} -1|^{k}}\\
\\
Now\ if\ k &gt;1:\ we\ notice\ that\lim _{n\rightarrow \infty }[ g( n)]^{k-1} =0\ \\
hence\ both\lim _{n\rightarrow \infty } |\ [ g( n)]^{k-1} *[ A+f( n)] *( -1)^{S( n)} -1\ |=|-1|=1\\
and\ \lim _{n\rightarrow \infty } |\ [ g( n-1)]^{k-1} *[ A+f( n-1)] *( -1)^{S( n-1)} -1|^{k} =1\\
and\ so\ LIMT=A*\frac{1}{1} =A\ so\ we\ are\ done.\\
\\
If\ k=1:\\
Then\ it\ is\ false:\\
Take\\
x_{n} =g( n) =2^{-n} *( -1)^{S( n)}\\
where\\
S( n) =\ repeating\ sequence\ of\ digits:\\
\{0,0,1\} \ \ \ \ \\
which\ repeats\ \ n\bmod 3\\
\lim _{n\rightarrow \infty }\frac{|g( n+1) |}{|g( n) |} =1/2\\
\\
Limit=\lim _{n\rightarrow \infty }\frac{|g( n+1) -g( n) |}{|g( n) -g( n-1) |} =\lim _{n\rightarrow \infty }\frac{|2^{-n-1} *( -1)^{S( n+1)} -2^{-n} *( -1)^{S( n)} |}{|2^{-n} *( -1)^{S( n)} -2^{-n+1} *( -1)^{S( n-1)} |}\\
\\
=\lim _{n\rightarrow \infty }\frac{|( -1)^{S( n+1)} -2( -1)^{S( n)} |}{|4( -1)^{S( n-1)} -2( -1)^{S( n)} |}\\
and\ this\ alternates\ between\ \frac{3}{2} ,\frac{1}{2} ,\frac{1}{6} \ and\ does\ not\ even\ appraoch\ a\ limit....\\
Unless\ you\ ask\ that\ the\ limit\ \lim _{n\rightarrow \infty }\frac{|x_{n+1} -x_{n} |}{|x_{n} -x_{n-1} |^{k}} MUST\ EXIST...\\
\\
\end{array}
\begin{array}{l}
If\ \lim _{n\rightarrow \infty } x_{n} =L\ \ for\ some\ L\in R,\ then\ we\ let\ g( n) =x_{n} -L\ hence\ \lim _{n\rightarrow \infty } \ g( n) =0\\
\\
\\
If\ \lim _{n\rightarrow \infty } x_{n} =L\ \ for\ some\ L\in R,\\
then\ we\ can\ conclude\ that\ x_{n} -L=g( n) \ for\ some\ g\ where\ \lim _{n\rightarrow \infty } g( n) =0\ \\
the\ first\ limit,\lim _{n\rightarrow \infty }\frac{|x_{n+1} -L|}{|x_{n} -L|^{k}} =A\ for\ some\ A\in R\\
then\ becomes\lim _{n\rightarrow \infty }\frac{|g( n+1) |}{|g( n) |^{k}} =A\ \\
we\ conclude\ that\ \frac{|g( n+1) |}{|g( n) |^{k}} -A=f( n) \ for\ some\ f\ where\ \lim _{n\rightarrow \infty } f( n) =0\ \\
hence\ g( n+1) =[ g( n)]^{k} *[ A+f( n)] *( -1)^{S( n)} \ where\ S\ is\ some\ function\ N\rightarrow N\\
We\ need\ the\ ( -1)^{S( n)} \ function\ because\ we\ dont\ know\ how\ the\ signs\ of\ these\ two\\
sides\ relate\ to\ one\ another\ but\ we\ know\ they\ are\ equal\ in\ magintude.\\
\\
So\ now\ to\ look\ at\ the\ second\ limit:\\
LIMIT=\lim _{n\rightarrow \infty }\frac{|x_{n+1} -x_{n} |}{|x_{n} -x_{n-1} |^{k}} =\lim _{n\rightarrow \infty }\frac{|g( n+1) -g( n) |}{|g( n) -g( n-1) |^{k}}\\
=\lim _{n\rightarrow \infty }\frac{|[ g( n)]^{k} *[ A+f( n)] *( -1)^{S( n)} -g( n) |}{|[ g( n-1)]^{k} *[ A+f( n-1)] *( -1)^{S( n-1)} -g( n-1) |^{k}}\\
\\
factoring\ out\ we\ get:\\
=\lim _{n\rightarrow \infty }\frac{|[ g( n)] |*|\ [ g( n)]^{k-1} *[ A+f( n)] *( -1)^{S( n)} -1\ |}{|[ g( n-1)]^{k} |*|\ [ g( n-1)]^{k-1} *[ A+f( n-1)] *( -1)^{S( n-1)} -1|^{k}}\\
\\
but\ we\ know\ \lim _{n\rightarrow \infty }\frac{|g( n+1) |}{|g( n) |^{k}} =A\ \ hence\ \lim _{n\rightarrow \infty }\frac{|g( n) |}{|g( n-1) |^{k}} =A\ and\ so\\
\\
LIMIT=A*\lim _{n\rightarrow \infty }\frac{|\ [ g( n)]^{k-1} *[ A+f( n)] *( -1)^{S( n)} -1\ |}{|\ [ g( n-1)]^{k-1} *[ A+f( n-1)] *( -1)^{S( n-1)} -1|^{k}}\\
\\
Now\ if\ k &gt;1:\ we\ notice\ that\lim _{n\rightarrow \infty }[ g( n)]^{k-1} =0\ \\
hence\ both\lim _{n\rightarrow \infty } |\ [ g( n)]^{k-1} *[ A+f( n)] *( -1)^{S( n)} -1\ |=|-1|=1\\
and\ \lim _{n\rightarrow \infty } |\ [ g( n-1)]^{k-1} *[ A+f( n-1)] *( -1)^{S( n-1)} -1|^{k} =1\\
and\ so\ LIMT=A*\frac{1}{1} =A\ so\ we\ are\ done.\\
\\
If\ k=1:\\
Then\ it\ is\ false:\\
Take\\
x_{n} =g( n) =2^{-n} *( -1)^{S( n)}\\
where\\
S( n) =\ repeating\ sequence\ of\ digits:\\
\{0,0,1\} \ \ \ \ \\
which\ repeats\ \ n\bmod 3\\
\lim _{n\rightarrow \infty }\frac{|g( n+1) |}{|g( n) |} =1/2\\
\\
Limit=\lim _{n\rightarrow \infty }\frac{|g( n+1) -g( n) |}{|g( n) -g( n-1) |} =\lim _{n\rightarrow \infty }\frac{|2^{-n-1} *( -1)^{S( n+1)} -2^{-n} *( -1)^{S( n)} |}{|2^{-n} *( -1)^{S( n)} -2^{-n+1} *( -1)^{S( n-1)} |}\\
\\
=\lim _{n\rightarrow \infty }\frac{|( -1)^{S( n+1)} -2( -1)^{S( n)} |}{|4( -1)^{S( n-1)} -2( -1)^{S( n)} |}\\
and\ this\ alternates\ between\ \frac{3}{2} ,\frac{1}{2} ,\frac{1}{6} \ and\ does\ not\ even\ appraoch\ a\ limit....\\
Unless\ you\ ask\ that\ the\ limit\ \lim _{n\rightarrow \infty }\frac{|x_{n+1} -x_{n} |}{|x_{n} -x_{n-1} |^{k}} MUST\ EXIST...\\
\\
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