1. [SOLVED] Series

Let $x_n$ be the sequence with $x_1=2$ and $x_n=\sqrt{5(x_{n-1})+6}$ for every integer $n\geq 2$. How can this be written in summation notation?

$x_1=2$
$x_2=\sqrt{16}=4$
$x_3=\sqrt{26}$
$x_4=\sqrt{5\big(\sqrt{26}\big)+6}$
$x_5=\sqrt{5\bigg(\sqrt{5\big(\sqrt{26}\big)+6}\big g)+6}$
......

2. Originally Posted by dwsmith
Let $x_n$ be the sequence with $x_1=2$ and $x_n=\sqrt{5(x_{n-1})+6}$ for every integer $n\geq 2$. How can this be written in summation notation?

$x_1=2$
$x_2=\sqrt{16}=4$
$x_3=\sqrt{26}$
$x_4=\sqrt{5\big(\sqrt{26}\big)+6}$
$x_5=\sqrt{5\bigg(\sqrt{5\big(\sqrt{26}\big)+6}\big g)+6}$
......
You shouldn't be writing it with summation notation unless you intend to sum the terms of the sequence.

3. Originally Posted by Prove It
You shouldn't be writing it with summation notation unless you intend to sum the terms of the sequence.
That is what I intend to do.

4. Originally Posted by dwsmith
That is what I intend to do.
What sum are you talking about? The $x_n$ terms aren't sums at all: they are, obviously, square roots. And if you are talking about $\sum_{n=1}^\infty x_n$ it is obvious that, since $\lim_{n\to \infty}x_n=6$, you have that $\sum_{n=1}^\infty x_n=+\infty$.

Maybe what you are really interested in is $\lim_{n\to\infty}x_n$?

5. Originally Posted by Failure
What sum are you talking about? The $x_n$ terms aren't sums at all: they are, obviously, square roots. And if you are talking about $\sum_{n=1}^\infty x_n$ it is obvious that, since $\lim_{n\to \infty}x_n=6$, you have that $\sum_{n=1}^\infty x_n=+\infty$.

Maybe what you are really interested in is $\lim_{n\to\infty}x_n$?
Why is the limit 6?

6. Originally Posted by dwsmith
Why is the limit 6?
Well, if the sequence $x_n := \sqrt{5x_{n-1}+6}$ does have a limit, $x_\infty$, say, then, by continuity we should have that

$x_\infty = \lim_{n\to \infty}x_n=\lim_{n\to\infty}\sqrt{5x_{n-1}+6}=\sqrt{5\lim_{n\to\infty}x_{n-1}+6}=\sqrt{5x_\infty+6}$

But the equation $x_\infty=\sqrt{5x_\infty+6}$ has only one solution, namely $x_\infty=6$.

Of course, this does not prove that the sequence, starting with $x_1=2$, actually does converge to $6$, but it shows that if the sequence converges at all, it converges to $6$.

That the sequence really does converge follows, for example, from Banch's Fixed Point Theorem (aka. Contraction Mapping Theorem), since $f:\; [4;15]\to [4;15], x\mapsto \sqrt{5x+6}$ is contracting.