let $x_n=\sqrt{1+\sqrt{2+\sqrt{3+....+\sqrt{n}}}}$ prove that $x_{n+1}^2 \leq 1+\sqrt{2}x_n$ and $x_n$ converges find an equivalent to $x_n$
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actually , i am not able to find the equivalent – Mike Sep 03 '19 at 21:08
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What is meant by "an equivalent?" – Charles Hudgins Sep 03 '19 at 21:15
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4Essentially a duplicate of How can I show that $\sqrt{1+\sqrt{2+\sqrt{3+\sqrt\ldots}}}$ exists? – mrtaurho Sep 03 '19 at 21:37
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To prove the inequality you just square $x_{n+1}$ like that:
$x_{n+1}^2=1+\sqrt{2+\sqrt{3+\dots \sqrt{n+1}}}$.
Now you use the simple fact, that
$\sqrt{2}\sqrt{1+\sqrt{2+\dots}}=\sqrt{2+2\sqrt{2\dots}}=\sqrt{2+\sqrt{8+\dots}}$
and you derive at the inequality.
You should take a look at How can I show that $\sqrt{1+\sqrt{2+\sqrt{3+\sqrt\ldots}}}$ exists?
