Let $x_0 = 1, x_{n+1} = x_n + \frac{1}{x_n}$; let $y_n = x_n^2 - 2n$ , show that $y_n$ is monotone increasing and $y_n < \log n + 2$

Asked Oct 06 '21 at 23:40

Active Oct 06 '21 at 23:40

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For the question as in the title, I have found that the sequence $x_n$ diverges, and I have come up with a recursive formula for $y_n$ as $y_{n+1} = y_n + \frac{1}{y_n + 2n}$; but I am struggling to find a way to prove $y_n < 2 + \log n$, and finally the question requires to prove that $\lim_{n \to \infty}x_n - \sqrt{2n} = 0$, which I think will depend on the previous inequality I'm struggling to prove.

Can anyone offer some hint on how I should proceed?

asked Oct 06 '21 at 23:40

Mar1129

1

Related – Arctic Char Oct 06 '21 at 23:53
Have you tried induction? – Vishu Oct 06 '21 at 23:53
Also this – Arctic Char Oct 06 '21 at 23:59
I attempted induction at first, but the $2n$ term in the denominator of the recurrence relation seems to make things more complicated... – Mar1129 Oct 07 '21 at 00:56
@ArcticChar Thanks the link really helps! Now I've figured it out! – Mar1129 Oct 07 '21 at 12:48

Let $x_0 = 1, x_{n+1} = x_n + \frac{1}{x_n}$; let $y_n = x_n^2 - 2n$ , show that $y_n$ is monotone increasing and $y_n < \log n + 2$

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