# Prove limit of recurrence relation

• Apr 1st 2007, 12:09 PM
slowcurv99
Prove limit of recurrence relation
Let S1 = sqrt(6), S2 = sqrt( 6 + sqrt(6) ), S3 = sqrt( 6 + sqrt(6) +sqrt(6) ), and in general define S(n+1) = sqrt( 6 + Sn ). Prove that Sn converges, and find its limit
• Apr 1st 2007, 12:26 PM
ThePerfectHacker
Quote:

Originally Posted by slowcurv99
Let S1 = sqrt(6), S2 = sqrt( 6 + sqrt(6) ), S3 = sqrt( 6 + sqrt(6) +sqrt(6) ), and in general define S(n+1) = sqrt( 6 + Sn ). Prove that Sn converges, and find its limit

Hint: Show it is a bounded monotone sequence.

Hint2: lim s_{n+1} = lim s_n
• Apr 1st 2007, 12:57 PM
Soroban
Hello, slowcurv99!

Quote:

Let S1 = sqrt(6), S2 = sqrt[6 + sqrt(6], S3 = sqrt{6 + sqrt[6 + sqrt(6)]},

and in general define: .S
n+1 .= .sqrt(6 + Sn).

Prove that S
n converges, and find its limit.

We have: .S .= .sqrt
{6 + sqrt[6 + sqrt(6) + ... ]}
. . . . . . . . . . . . . . . . . . .\_________________/
. . . . . . . . . . . . . . . . . . . . . . .
This is S
. . . . . . . . . . . . . . . . . . ____
Hence, we have: . S .= .√6 + S

Square both sides: . .= .6 + S . . Sē - S - 6 .= .0

. . which factors: .(S - 3)(S + 2) .= .0

. . and has roots: .S = 3, -2

Taking the positive root: .S = 3