Determine if the sequence converges

Question

For a sequence $\{x_n\}$ given by $x_1=3$, $x_n= \frac{x_{n-1}+2}{x_{n-1}}$ how would I determine if this sequence converges since it is not monotonic?

Answer 1

Hint: the sequence is positive, and$\;x_n=1+\dfrac{2}{x_{n-1}}\,$, so $\,x_n-x_{n-1}=\dfrac{-2(x_{n-1}-x_{n-2})}{x_{n-1}x_{n-2}}\,$, which means that the difference between consecutive terms changes sign at each step, and therefore the subsequences of odd, respectively even, indices are monotonic.

Answer 2

Yes, it converges to $2$

Note that the subsequence $$\{ x_{2n-1}\}$$ is decreasing and the subsequence $$\{ x_{2n}\}$$ is increasing. We have a nested interval situation where these two subsequence converge to a common limit.

The limit satisfies $$ L= \frac {L+2}{L}$$

Which gives us two candidates $2$ and $-1$

Of course we pick the positive candidate $2$

Answer 3

Simplifying the expression:

$$ x_n = 1+\frac{2}{x_{n-1}} $$

We note that for positive $x_1$, the series is always positive and always larger than 1. So, if there was ever a limit $L$, we could find it by:

$$ L = 1 +\frac{2}{L} $$ $$ L^2 = L +2 $$ $$ L_1=-1 \, \text{and}\,L_2=2 $$

Since the first solution is impossible, the question is if starting with 3, the sequence eventually reaches 2, which is a fixed point. Note that:

$$ f(x)=1+\frac{2}{x} $$

Implies: $$ f'(x)=-\frac{2}{x^2} $$

And I note that for $x>\sqrt{2}$ we ha have that $|f'(x)|<1$, hence $f$ is a contraction, therefore its iteration converges to a fixed point. And as seen above, doing the iteration after starting with 3 will never cause us to have $x\leq \sqrt{2}$.

Answer 4

We will show the limit is $2$. Write down the first few terms and notice the continued fraction: $$x_{100}=1+\frac{2}{1+\frac{2}{1+\frac{2}{1+\frac{2}{\vdots} }}}.$$ Hence: $$\begin{align}2=1+1=&1+\frac{2}{2}=\\ &1+\frac{2}{1+\frac22}=\\ &1+\frac{2}{1+\frac{2}{1+\frac{2}{\vdots} } } \end{align}.$$

Answer 5

I shall show that the limit is $2$. Note that $x_k>1 $ for all $k$ and also

$$x_{n+1}-2 = \frac{2}{x_n}-1= -\frac{x_n-2}{x_n}\tag{1}$$ This fact tells us that if $x_k>2$, then $x_{k+1}<2$ and vice-versa. Also, $$x_{2n}-2= \frac{x_{2(n-1)}-2}{x_nx_{n-1}}\tag{2}$$ Using the above statement and relation, along with the fact that $x_2 = 5/3<2$, we have that $x_k>x_2=5/3$ for all $k$. Thus, using relation $(1)$ we have: $$|x_{n+1}-2|=|\frac{x_n-2}{x_n}|<\frac{3}{5}|x_n-2|$$

Thus we have $|x_{n+1}-2|\le (\frac{3}{5})^{n-2}|x_2-2| \to 0 \text{ as }n\to\infty$. Thus the limit of $x_n$ tends to $2$

Answer 6

Suppose $(x_n)$ converges to $x$. Then $\lim(x_n)=\lim{\frac{x_{n-1}+2}{x_{n-1}}}$

Using algebra of limits, we have that $x=\frac{x+2}{x}$

$\implies x^2-x-2=0$

Using the quadratic formula, we get $x=2$ or $ x=-1$

Since $x_n \geq 0 \forall n \in \mathbb{N}$, this implies that the only value $(x_n)$ can converge to is $2$

Now, we need to see the behaviour of the terms of the sequence.

$x_1=3$, $x_2 = 5/3$ $x_3=11/5$...

You'll notice that the odd subsequence $(x_{2n-1})$ is decreasing while the even subsequence $(x_{2n})$ is increasing. Also, note that $x_{2n} \leq x_{2n-1}$ for all $n \in \mathbb{N}$

And so, we have a situation like this:

$0 \leq x_{2n} \leq x_{2n-1} \leq 3$

Then, both the even and odd subsequences are bounded and since they are monotone, they converge to their supremum and infimum respectively. Then, both these subsequences will converge to the same value, and thus, the whole sequence will converge to 2.

Note: There is a theorem that says that if both the even and odd subsequences of a sequence converge to the same value, then the sequence converges to that value.

Answer 7

The function $$T(x):={x+2\over x}$$ can be regarded as a Moebius transformation. It has the fixed points $-1$ and $2$. We therefore introduce a new complex coordinate $$u:={x-2\over x+1}$$ on the Riemann sphere. Thereby the point $x=2$ corresponds to $u=0$, and the point $x=-1$ to $u=\infty$. In terms of this coordinate the given $T$ appears (after some calculation) as $$\hat T(u)=-{u\over2}\ .$$ This reveils the point $u=0$ as an attractive, and $u=\infty$ as a repelling fixed point of $\hat T$. In terms of $T$ this means that the sequence $(x_n)_{n\geq0}$ converges to $2$ for any initial point $x_0\ne-1$.