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
Hello, slowcurv99!
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