Question #7838

Let x_sub(1) = a>0 and x_sub(n+1) = x_sub(n) + 1/x_sub(n). Show that x_sub(n) diverges

Expert's answer

Question 1. Let x1=a>0x_{1} = a > 0 and xn+1=xn+1/xnx_{n + 1} = x_{n} + 1 / x_{n}. Show that xnx_{n} diverges.

Solution. Suppose there is b=limxn<b = \lim x_{n} < \infty. Note that b>0b > 0, because x1>0x_{1} > 0 and (xn)(x_{n}) is increasing. Then consider the equality xn+1=xn+1/xnx_{n + 1} = x_{n} + 1 / x_{n}. As nn \to \infty the left-hand side of this equality tends to bb, while the right-hand side tends to b+1/bb + 1 / b. Since bb+1/bb \neq b + 1 / b, we obtain a contradiction.

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