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

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- April 1st 2007, 12:09 PMslowcurv99Prove 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

- April 1st 2007, 12:26 PMThePerfectHacker
- April 1st 2007, 12:57 PMSoroban
Hello, slowcurv99!

Quote:

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

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

Prove that Sn 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: .Sē .= .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**