How to prove that the sequence $y_{n+1}=1-2y_n$ where $y_1=0$ has no convergent subsequences How to prove that the sequence $y_{n+1}=1-2y_n$ where $y_1=0$ has no convergent subsequences?
I don't know where to even begin with this proof. I'm assuming I need to prove that $|y_n|$ converges to infinity first.
 A: We have: $|y_{n+1}| = |2y_n - 1| \ge 2|y_n| - 1\ge 2(2|y_{n-1}| - 1)-1=4|y_{n-1}|-3\ge 4(2|y_{n-2}|-1)-3=8|y_{n-2}|-7\ge 8(2|y_{n-3}|-1)-7=16|y_{n-3}|-15 \ge 2^{n-3}|y_4|-(2^{n-3}-1)=3\cdot 2^{n-3}-2^{n-3}+1=2^{n-2}+1, \forall n\ge 4$. Thus $|y_{n}| \to +\infty$. You can fill the rest of the work.
A: The first few values are
$0, 1, -1, 3, -5,
11, -21
$.
So it looks like it is
growing exponentially
with alternating signs.
Any sequence with
$|y_{n+1}|
\gt |y_n|+a
$
for some $a > 0$
and all large enough $n$
can not have a convergent subsequence.
If it does
and the subsequence is
$(y_{n_k})$,
then,
for any $c > 0$
$|y_{n_{k+1}}-y_{n_k}|
\lt c
$
for all large enough $k$.
Since
$|y_{n+1}|
\gt |y_n|+a
$,
$|y_{n+k}|
\gt |y_n|+ka
$
so,
if $n > m$,
$|y_{n}-y_m|
\gt (n-m)a
$.
Therefore
$|y_{n_{k+1}}-y_{n_k}|
\gt a(n_{k+1}-n_k)
\ge a$.
This implies that
$c > a$
which is false
as we can chose $0 < c < a$.
In our case,
if $|y_n| > 2$
then
$|y_{n+1}|
=|1-2y_n|
\ge 2|y_n|-1
$
so
$|y_{n+1}|-|y_n|
\ge |y_n|-1
\gt 1$
and
$|y_{n+1}|
\gt 2$.
Therefore
$y_n$
does not have a
convergent subsequence.
We can also find
an explicit form.
In this case,
based on experience,
I would try to find a $c$
such that
$y_{n+1}+c
=-2(y_n+c)
$.
From this,
an explicit form is
readily found.
Regrouping this,
$y_{n+1}
=-3c-2y_n
$,
so $-3c=1$
or
$c=-\frac13$.
As a check,
if
$y_{n+1}-\frac13
=-2(y_n-\frac13)$,
then
$y_{n+1}
=1-2y_n
$.
By induction,
$y_{n}-\frac13
=(-2)^k(y_{n-k}-\frac13)$.
Setting $k=n-1$,
$y_{n}-\frac13
=(-2)^{n-1}(y_{1}-\frac13)
=-\frac13(-2)^{n-1}
$
so
$y_{n}
=\frac13(1-(-2)^{n-1})
$.
A: This is a linear recurrence for which you can have an expression for the general term. You can start by solving the homogeneous equation, $y_{n+1} = -2 y_n$, yielding $y_n^h = c (-2)^n$ and then find a particular constant solution. This way you conclude that $y_n = c (-2)^n + \frac 13$. Finally, using the initial condition, $y_n = \frac 16 (-2)^n + \frac 13$. Given the general term for this sequence, the conclusion follows trivially.
A: $y_{n+1}=1-2y_n, y_1=0$. $0,1,-1,3,-5,11,-21,43:1,-2,4,-8,16,-32,64$
$y_n=\frac{(-2)^{n-1}-1}{-3}$. Take consecutive differences and one finds the original sequence is a the geometric series starting with 1 and having common ratio -2. This tells us all the elements of the sequence are distinct integers. The smallest possible difference between two distinct integers is 1. One can use this to prove there are no convergent subsequences.
Suppose some subsequence of the set converges to real  $a$. Then by definition, $\forall \epsilon>0 \exists \ y_n \ s.t. |y_n-a|<\epsilon$
$a-\epsilon<y_n<a+\epsilon$. We must have some $y_m \ s.t. y_n<y_m<a+\epsilon $ else we can select $\epsilon=(a-y_n)/2$ and disprove convergence. On the other hand. $|y_n-a|<|y_m-a|<\epsilon$
$|y_m-y_n|=|y_m-a + a-y_n|\le |y_m-a|+|y_n-a|<2\epsilon$. We can choose any $\epsilon$ because of convergence. Pick $\epsilon=1/4$ and we have $|y_m-y_n|<1/2$, two integers whose difference is less than 1. Another contradiction.
