We can easily see that (x_n) is the decreasing sequence.

Because by A-G inquality, we have:

(x_n) + 1/(x_n) > 2, then (x_n) + 1/(x_n) - 2 > 0, so x_(n+1) - x_n = 2 - 1/(x_n) - (x_n) < 0

It means that x_(n+1) < x_n. The decreasing sequence.

Otherwise, we prove that x_n >= 1 for all n in N.

It is easy to use reduction method. ^_^

From above, (x_n) converges.

Assume that limx_n = a. Then a = 2 - 1/a

i.e a = 1.

In conclusion, limx_n = 1.

enable f(n)(x) = a million + x +x2 + x3 + ... +xn, and g(n)(x) = (xn+a million -a million)/(x-a million) a million. enable us to analyze no count if f(a million)(x) =?= g(a million)(x) f(a million)(x) = a million + x g(a million)(x) = (x^2 - a million) / (x - a million) = x + a million = f(a million)(x) 2. enable us to assume that for some m, that f(m)(x) = g(m)(x) f(m+a million)(x) = f(m) + x^(m + a million) g(m+a million)(x) = (x^(m+2) - a million) / (x - a million) = (x^(m+2) - x^(m+a million) + x^(m+a million) - a million) / (x - a million) = (x^(m+2) - x^(m+a million)) / (x - a million) + (x^(m+a million) - a million) / (x - a million) = x^(m+a million) + g(m)(x) = x^(m+a million) + f(m)(x) = f(m+a million)(x) From a million. & 2., it particularly is been shown, by technique of mathematical induction, that f(n)(x) = g(n)(x) for all organic n.