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Compute the limit:

$$\lim_{n \to \infty} n^2\sqrt{1-\cos(1/n)+\sqrt{1-\cos(1/n)+\sqrt{1-\cos(1/n)+\ldots}}}$$

Did
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Simar
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    Are there $n$ radicals or infinite radicals? – Kunnysan Jul 26 '13 at 22:19
  • @Kunnysan :$\dots$ means continued indefinitely, for anything to mean 'n' terms usage,then after $dots$ a termwill be placed to signify that is a terminating sequence. – jimjim Jul 26 '13 at 22:29
  • @Simar : the correct term is indefinite not infinite. – jimjim Jul 26 '13 at 22:30
  • @Simar : There is no doubt, $\dots$ mean indefinite, do not change the question when people need to learn the meaning of well used notation,teach them what is correct. – jimjim Jul 26 '13 at 22:33
  • @Arjang Why are you after that indefinate/infinite sign? Whats wrong in putting that? – Simar Jul 26 '13 at 22:33
  • @Simar : It is not mathematics! Imagine it was an English sentence that was correct to begin with then someboy kept adding something that is not needed or correct to be used, for example instead of "Jack climbed the tree" , to make sure the tree was only one tree change it to "jack climbed the a tree", it is the same with using $\infty$ when it is not meant to be there. PS: I used to do same thing too, till I learned better. – jimjim Jul 26 '13 at 22:39

2 Answers2

5

Hints:

  • For every $a\gt0$, $b=\sqrt{a+\sqrt{a+\sqrt{a+\cdots}}}$ is such that $b^2+a=b$ and $b\gt0$, thus $b=\frac12+\frac12\sqrt{1+4a}$.
  • When $n\to\infty$, $1-\cos(1/n)\to0$.
  • When $a\to0$, $\frac12+\frac12\sqrt{1+4a}\to1$.
  • Hence the limit you are after is $\lim\limits_{n\to\infty}n^2\cdot1=+\infty$.
Did
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Define $$ a_m=\underbrace{\small\sqrt{1-\cos(1/n)+\sqrt{1-\cos(1/n)+\sqrt{1-\cos(1/n)+\ldots+\sqrt{1-\cos(1/n)}}}}}_{m\text{ square roots}} $$ $a_1=\sqrt{1-\cos(1/n)}=\sqrt{2\sin^2\left(\frac1{2n}\right)}\lt2$. If $a_m\le2$, then $$ \begin{align} a_{m+1} &=\sqrt{1-\cos(1/n)+a_m}\\ &\le\sqrt{2+2}\\[4pt] &=2 \end{align} $$ Furthermore, note that $a_{m+1}\ge a_m$. Thus, $a_m$ is an increasing sequence, bounded above; therefore, $$ A_n=\lim_{m\to\infty}a_m $$ exists, is non-negative, and is no greater than $2$. We then have $$ A_n^2=2\sin^2\left(\frac1{2n}\right)+A_n $$ Solving for $A_n$ yields $$ A_n=\frac{1+\sqrt{1+8\sin^2\left(\frac1{2n}\right)}}{2} $$ $A_n\ge1$ for all $n$, therefore, $$ \lim_{n\to\infty}n^2A_n=\infty $$

robjohn
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