My first term at $n=1$ is $$\sqrt 2 $$ 2nd term at $n=2$ is $$\sqrt {\sqrt 2+2} $$ So I am adding 2 to the previous term then taking the square root of the entire equation. At $n=3$ I have $$\sqrt {\sqrt {\sqrt {2}+2}+2}$$ And so on. I want to sum all terms to infinite.
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2The terms actually approaches to $2$. It can be solved by the equation $x^2-2=x$ . So the sum would not converge since you are adding more and more – lEm Apr 15 '16 at 12:17
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2$x_1 = \sqrt{2}$, $x_{n+1} = \sqrt{x_n + 2}$ – Edward Evans Apr 15 '16 at 12:18
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1Interestingly, each term in your sequence has the closed form $2\cos \frac{\pi}{2^n}$ – zz20s Apr 15 '16 at 14:10
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@zz20s cool! How to prove that? – Ant Apr 15 '16 at 23:16
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@ant You may want to do a google search for "Viete's formula for pi." – zz20s Apr 15 '16 at 23:18
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So you have $a_1 = \sqrt 2$, $a_n = \sqrt{a_{n-1} + 2}$
Now suppose that $\lim_{n \to \infty} a_n = S$ exists, then it must hold
$$S = \sqrt{S + 2}$$ which has as positive solution $S = 2$
Now, you want to find $$\sum_{n =1}^\infty a_n$$
A necessary condition for the sum to converge is that $$\lim_{n \to \infty} a_n = 0$$ So either the limit does not exists (and the sum does not converge) or the limit exists and for what we showed before it must be $2 \neq 0$; hence your sum does not converge
Ant
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