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7.7 The Translation Theorem

7.73   Theorem. Let $f$ be a real convergent sequence, say $f\to L$. If $f(n)\geq 0$ for all $n\in\mbox{{\bf N}}$, then $L\geq 0$.


Proof: I note that $L\in\mbox{{\bf R}}$, since if $f\to L$, then $\mbox{\rm Re}f\to\mbox{\rm Re}L$. Suppose, to get a contradiction, that $L<0$, (so $\displaystyle {-{L\over 2}>0}$), and let $N_{f-\tilde L}$ be a precision function for the null sequence $f-\tilde L$. Let $\displaystyle {N=N_{f-\tilde
L}\left(-{L\over 2}\right)}$. Then $\displaystyle {\vert(f-\tilde L)(N)\vert\leq -{L\over 2}}$, so $\displaystyle {\vert f(N)-L\vert\leq -{L\over 2}}$, and hence $\displaystyle {f(N)<L-{L\over 2}={L\over 2}<0}$. This contradicts the assumption that $f(N)\geq 0$ for all $n\in\mbox{{\bf N}}$. $\mid\!\mid\!\mid$

7.74   Exercise (Inequality theorem.) A Let $f,g$ be convergent real sequences. Suppose that $f(n)\leq g(n)$ for all $n\in\mbox{{\bf N}}$. Prove that $\lim f\leq\lim g$.

7.75   Exercise. A Prove the following assertion, or give an example to show that it is not true. Let $f,g$ be convergent real sequences. Suppose that $f(n)<g(n)$ for all $n\in\mbox{{\bf N}}$. Then $\lim f<\lim g$.

7.76   Definition (Translate of a sequence.) Let $f$ be a sequence and let $p\in\mbox{{\bf N}}$. Then the sequence $f_p\colon n\mapsto
f(n+p)$ is called a translate of $f$.

7.77   Example. If $\displaystyle {f=\left\{ {1\over {2^2}},{1\over {3^2}},{1\over {4^2}},\cdots,{1\over
{(n+2)^2}},\cdots\right\}}$, then $\displaystyle {f_3=\left\{ {1\over {5^2}},{1\over
{6^2}},{1\over {7^2}},\cdots,{1\over {(n+5)^2}},\cdots\right\}}$. A translate of a sequence is a sequence obtained by ignoring the first few terms.

7.78   Theorem (Translation theorem.) If $\{f(n)\}$ is a convergent complex sequence, and $p\in\mbox{{\bf N}}$, then $\{f(n+p)\}$ converges, and $\lim\{f(n)\}=\lim\{f(n+p)\}$. Conversely, if $\{f(n+p)\}$ converges, then $\{f(n)\}$ converges to the same limit.


Proof: Let $f\to L$, let $f_p(n)=f(n+p)$ and let $N_{f-\tilde L}$ be a precision function for $f-\tilde L$. I claim $N_{f-\tilde L}$ is also a precision function for $f_p-\tilde L$. In fact, for all $n\in\mbox{{\bf N}}$, and all $\varepsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$,

\begin{displaymath}n\geq N_{f-\tilde L}(\varepsilon)\mbox{$\hspace{1ex}\Longrigh...
...x}\Longrightarrow\hspace{1ex}$}\vert f(n+p)-L\vert<\varepsilon.\end{displaymath}

Conversely, suppose

\begin{displaymath}\{f_p(n)\}=\{f(n+p)\}\to L\end{displaymath}

and let $N_{f_p-\tilde L}$ be a precision function for $f_p-\tilde L$. Let $N(\varepsilon)=p + N_{f_p-\tilde L}(\varepsilon)$ for all $\varepsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$. I claim $N$ is a precision function for $N_{f-\tilde L}$. For all $n\in\mbox{{\bf N}}$,

\begin{eqnarray*}
n>N(\varepsilon)&\mbox{$\Longrightarrow$}&n>p+N_{f_p-\tilde L}...
...arrow$}&\vert f(n)-L\vert<\varepsilon.\mbox{ $\mid\!\mid\!\mid$}
\end{eqnarray*}



7.79   Example. Let the sequence $f$ be defined by

\begin{eqnarray*}
f(0)&=&1, \\
f(n+1)&=&{1\over {1+f(n)}} \mbox{ for all } n\in\mbox{{\bf N}}.
\end{eqnarray*}



Then

\begin{eqnarray*}
f(1)&=&{1\over {1+1}}={1\over 2} \\
f(2)&=&{1\over {1+{1\over...
...1\over {1+{1\over {1+1}}}}}}={1\over {1+{2\over 3}}}={3\over 5}.
\end{eqnarray*}



Suppose I knew that $f$ converged to a limit $L$. It is clear that $f(n)>0$ for all $n$, so $L$ must be $\geq 0$. By the translation theorem

\begin{displaymath}L=\lim\{f(n+1)\}=\lim\left\{ {1\over {1+f(n)}}\right\}={1\over {1+\lim
f(n)}}={1\over {1+L}}\end{displaymath}

so $L(1+L)=1$; i.e., $L^2+L-1=0$. Hence $\displaystyle { L\in\left\{ {{-1+\sqrt{1+4}}\over
2},\; {{-1-\sqrt{1+4}}\over 2}\right\}}$, and since $L\geq 0$, we conclude $\displaystyle {L={{\sqrt
5-1}\over 2}}$. I've shown that the only thing that $f$ can possibly converge to is $\displaystyle { {{\sqrt 5-1}\over 2}}$. Now

\begin{displaymath}0 < L < {3-1\over 2} = 1, \mbox{ so } \vert 1 - L \vert < 1.\end{displaymath}

Since $L = {1\over 1+L}$, we have for all $n\in\mbox{{\bf N}}$,

\begin{eqnarray*}
\vert f(n+1)- L\vert &=& \left\vert {1\over 1+f(n)} - {1\over ...
...rt \over 1+L}\\
&=& L\vert L- f(n)\vert = L\vert f(n) - L\vert.
\end{eqnarray*}



Hence

\begin{eqnarray*}
\vert f(1) - L\vert & \leq & L\vert f(0) - L\vert = L\vert 1-L...
...
\vert f(3) - L\vert & \leq & L\vert f(2) - L\vert \leq L^3,\\
\end{eqnarray*}



and by induction,

\begin{displaymath}\vert f(n)-L\vert \leq L^n \mbox{ for all }n \in \mbox{{\bf Z}}_{\geq 1}.\end{displaymath}

By theorem 7.64 $\{L^n\}$ is a null sequence, and by the comparison theorem for null sequences, it follows that $\{f(n) - L\}$ is a null sequence. This completes the proof that $f\to L$. $\mid\!\mid\!\mid$

7.80   Exercise. Let

\begin{eqnarray*}
f(0)&=&-2 \\
f(n+1)&=&{{f(n)^2+2}\over {2f(n)}} \mbox{ for all } n\in\mbox{{\bf N}}.
\end{eqnarray*}



a)
Assume that $f$ converges, and determine the value of $\lim\{f(n)\}$.
b)
Calculate $f(1),f(2),f(3),f(4)$, using all of the accuracy of your calculator. Does the sequence appear to converge?

7.81   Entertainment. Show that the sequence $f$ defined in the previous exercise converges. We will prove this result in Example 7.97, but you can prove it now, using results you know.

7.82   Exercise. Let $g$ be the sequence defined by

\begin{eqnarray*}
g(0) &=& 1,\\
g(1) &=& 1,\\
g(n+2) &=& {1+g(n+1) \over g(n)} \mbox{ for all }n \in\mbox{{\bf N}}.
\end{eqnarray*}



a)
Assume that $g$ converges, and determine the value of $\lim\{g(n)\}$.
b)
Calculate $g(1),g(2),g(3),g(4),g(5),g(6)$, using all of the accuracy of your calculator. Does this sequence converge?

7.83   Theorem (Divergence test.) Let $f,g$ be complex sequences such that $g(n)\neq 0$ for all $n\in\mbox{{\bf N}}$. Suppose that $g\to 0$ and $f\to L$ where $L\neq 0$. Then $\displaystyle {{f\over g}}$ diverges.


Proof: Suppose, to get a contradiction, that $\displaystyle {{f\over g}}$ converges to a limit $M$. Then by the product theorem, $\displaystyle {g\cdot{f\over g}}$ converges to $0\cdot M=0$; i.e., $f\to 0$. This contradicts our assumption that $f$ has a non-zero limit. $\mid\!\mid\!\mid$

7.84   Exercise. Prove the following assertion or give an example to show that it is not true: Let $f,g$ be complex sequences such that $g(n)\neq 0$ for all $n\in\mbox{{\bf N}}$, but $g\to 0$. Then $\displaystyle {{f\over g}}$ diverges.

7.85   Example. Let $\displaystyle {f(n)=\left\{ {{n^3+3n}\over {n^2+1}}\right\}}$ for all $n\in\mbox{{\bf Z}}_{\geq 1}$. Then

\begin{displaymath}f(n)={{n^3\left(1+{3\over {n^2}}\right)}\over {n^2\left(1+{1\...
...r {n^2}}\right)}\over { {1\over n} \left(1+{1\over n}\right)}}.\end{displaymath}

Since

\begin{displaymath}\lim\left\{\left(1+{3\over {n^2}}\right)\right\}_{n\geq 1}=1+0\neq 0,\end{displaymath}

and

\begin{displaymath}\lim\left\{{1\over n}\left(1+{1\over n}\right)\right\}_{n\geq 1}=0\cdot(1+0)=0,\end{displaymath}

it follows that $f$ diverges.

7.86   Exercise. A Let $A,B,a,b$ be complex numbers such that $an+b\neq 0$ for all $n\in\mbox{{\bf Z}}_{\geq 1}$. Discuss the convergence of $\displaystyle {\left\{ {{An+B}\over
{an+b}}\right\}_{n\geq 1}}$. Consider all possible choices for $A,B,a,b$.


next up previous index
Next: 7.8 Bounded Monotonic Sequences Up: 7. Complex Sequences Previous: 7.6 Geometric Series   Index