[SOLVED] Series

**dwsmith** · May 11th 2010, 02:24 PM

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?

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

**Prove It** · May 11th 2010, 04:38 PM

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.

**dwsmith** · May 11th 2010, 04:39 PM

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.

**Failure** · May 12th 2010, 07:26 AM

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 $\lim_{n\to\infty}x_n$ ?

**dwsmith** · May 12th 2010, 11:05 AM

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?

**Failure** · May 12th 2010, 08:05 PM

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.

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