Let S1 = sqrt of 6 and Sn+1 = sqrt(6 + Sn) for n>= 1; Prove that Sn converges and find it's limit

I know that I can use the ratio or root test to see if it works.

I believe it is montonic, but not too sure why.

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- Jun 24th 2006, 12:45 PMNichelle14How do I do this?
Let S1 = sqrt of 6 and Sn+1 = sqrt(6 + Sn) for n>= 1; Prove that Sn converges and find it's limit

I know that I can use the ratio or root test to see if it works.

I believe it is montonic, but not too sure why. - Jun 24th 2006, 04:37 PMSoroban
Hello, Nichelle!

We can find the limit with straight Algebra . . .

Quote:

14] Let $\displaystyle S_1 = \sqrt{6}$ and $\displaystyle S_{n+1} = \sqrt{6 + S_n}$ for $\displaystyle n \geq 1$

Prove that $\displaystyle S_n$ converges and find it's limit.

$\displaystyle S_1 \;= \;\sqrt{6}$

$\displaystyle S_2 \;= \;\sqrt{6 + \sqrt{6}}$

$\displaystyle S_3 \;= \;\sqrt{6 + \sqrt{6 + \sqrt{6}}} $

. $\displaystyle \vdots$ . . . . . . . . . $\displaystyle \vdots$

$\displaystyle S \;= \;\sqrt{6 + \underbrace{\sqrt{6 + \sqrt{6 + \sqrt{6 + \hdots}}}}} $

. . . . . . . . . . . . . . . This is $\displaystyle S$

So we have: .$\displaystyle S \;= \;\sqrt{6 + S}$

Square both sides: .$\displaystyle S^2 \;= \;6 + S$

We have a quadratic: .$\displaystyle S^2 - S - 6 \;= \;0$

. . which factors: .$\displaystyle (S - 3)(S + 2) \;= \;0$

. . and has the positive root: .$\displaystyle S \,= \,3$

Therefore: .$\displaystyle \lim_{n\to\infty} S_n \;=\;3$ . . . obviously, the sequence conveges.

Edit: Sorry ... I had a*very*stupid typo ... corrected now. - Jun 24th 2006, 05:51 PMNichelle14
How do you get 2 to be the postivie root?

You have (s-3)(s+2) = 0

so isn't this true

s-3 = 0 which is s = 3

and

s+2 = 0 which is s = -2

Just checking? - Jul 4th 2006, 09:39 AMRebesques
Soroban has got the method right! We just need some extra justification (that boring stuff pure mathematicians bother people with).

You can prove by induction, that the sequence is increasing and bounded above. By a well known theorem of calculus, the sequence converges. This justifies the step

$\displaystyle

S \;= \;\sqrt{6 + \underbrace{\sqrt{6 + \sqrt{6 + \sqrt{6 + \hdots}}}}}

$

Soroban used to calculate the actual limit :) . Now, you can see that all terms of the sequence remain positive; So the limit s=-2 is not an option. - Jul 4th 2006, 02:10 PMThePerfectHackerQuote:

Originally Posted by**Rebesques**

- Jul 5th 2006, 02:16 AMsrulikbd
can someone please find the formula for the nth number, and then we can just limit n>infinte, right?

- Jul 5th 2006, 04:23 AMRebesquesQuote:

can someone please find the formula for the nth number

Maybe we should just be happy we can tackle questions of convergence without seeing a recognizable pattern in the sequence. :cool: