For $a_1=\sqrt a , a_2=\sqrt{a-\sqrt{a}}, a_3=\sqrt{a-\sqrt{a+\sqrt {a}}} ,\ldots $ , We have $a_n$ as    $$\lim_{n\rightarrow\infty}a_n=\frac {A-1}6+\frac 23\sqrt{4a+A}\sin\left(\frac 13\arctan\frac {2A+1}{3\sqrt3}\right)\tag1$$ where $A=\sqrt{4a-7}$   Note: We have $- , + , +$ every three terms.  How we can prove this relation? (I want to prove it in general not only $a=2$ and period is $3$ not $4$)
124199
Supposedly, the infinitely nested radical $$\sqrt{a-\sqrt{a+\sqrt{a+\ldots}}}\tag1\label{1}$$ converges to $$\frac {A-1}{6}+\frac 23\sqrt{4a+A}\sin\left(\frac 13\arctan\frac {2A+1}{3\sqrt 3}\right)\tag2$$ where $A=\sqrt{4a-7}$. In fact, that's how Ramanujan arrived at the famous nested radical $$\sqrt{2-\sqrt{2+\sqrt{2+\ldots}}}=2\sin\left(\frac {\pi}{18}\right)\tag3$$ So I'm wondering how you would prove $\ref{1}$. My best try was to set $(1)$ equal to $x$ and substitute to get $$x=\sqrt{a-\sqrt{a+\sqrt{a+x}}}$$to get an octic polynomial. But I'm not too sure how the trigonometry made its way into the generalization.  And I also wonder if there is a way to generalize this even further to possible $$\sqrt{a-\sqrt{a-\sqrt{a+\sqrt{a+\ldots}}}}$$ which has period $4$ instead of $3$.
92934
If $a$ and $b$ are positive real numbers such that $a&lt;b$ and if $$a_1=\frac{a+b}{2}, b_1=\sqrt{(a_1b)},..., a_n=\frac{a_{n-1}+b_{n-1}}{2},b_n=\sqrt{a_nb_{n-1}},$$ then show that   $$\lim_{n\to\infty}b_n=\frac{\sqrt{b^2-a^2}}{\arccos\frac{a}{b}}.$$ I tried to calculate explicitly the first few terms $a_2,b_2$ etc but the terms got too complicated really quickly and I couldn't spot any pattern.
26450
I managed to show that the limit exists, but I don't know how to compute it.  EDIT:  There are initial terms: $a_1=1$ and $a_2=2$.
37001
Let $a=\frac{1+i\sqrt 7}{2}$ and $u_n=\Re(a^n)$   show that  $(|u_n|)\to +\infty$   I think basics method does not works here.  Any ideas ?
57583
Let $a,b,c&gt; 0$ be three real numbers such that $a+b+c=1$. I want to prove that   $$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+a^3+c^3}+\frac{c}{c^2+a^3+b^3}\le\frac{1}{5abc}.$$  My attempt: Using AM-GM on each denominator gives (here, LHS denotes the left-hand side) $$LHS\le\frac{1}{3a^{\frac12}bc}+\frac{1}{3b^{\frac12}ac}+\frac{1}{3c^{\frac12}ab}.$$  However, I think that my attempt doesn't work because the last expression can get larger than $\frac{1}{5abc}$. In fact, if we multiply with $3abc$ then the original inequality is: $$\sqrt a+\sqrt b+\sqrt c\le\frac35$$  However, as $a,b,c&lt;1$, $$\sqrt a +\sqrt b + \sqrt c&gt; a+b+c=1&gt;\frac35.$$ So my upper bound is always larger than the given one.
103459
Suppose $f:\mathbb{R}\to\mathbb{R}$ has period 1, and for some $q\in(0,1)$:  $$|f(x)-f(y)|\leq q|x-y|\quad \forall x,y$$  Let $g(x)=x+f(x)$, for any $a\in\mathbb{R}$,define the following sequence:  $$a_1=a,\quad a_2=g(a_1),\quad a_3=g(a_2),\quad \dots,\quad a_{n+1}=g(a_n)$$  Show that  $$\lim_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{n^2}$$  exists and does not depend on the choice of $a$.  I don't even know what is the possible approach, any hints? Thanks!
109854
I am stuck at   $$(a+d)^2 - 4(ad-bc) &lt; 0$$ $$\implies (a+d)^2&lt;4(ad-bc)$$ $$\implies (a+d)&lt;2\sqrt{ad-bc}$$  where $a,b,c,d \ge 0$.  Is there a contradiction to derive here?   Also, the square root on the R.H.S. may not be legit, since I don't have any more information about the relation of $ad$ and $bc$ -- so let's say I took the complex square root, to be safe.    (This lack of information also makes it hard to use the AM-GM inequality.)  Any help is appreciated.  Thanks,
33545
If $a_1,a_2,a_3,...,a_n$ are the side lengths of $A_1A_2A_3...A_n$ convex polygon,then$\frac{a^2_1+a^2_2+a^2_3+....+a^2_{n-1}}{a^2_n}$ is   $(A)&gt;\frac{1}{n-1}\hspace{1cm}(B)&lt;\frac{1}{n-1}\hspace{1cm}(C)&gt;\frac{1}{n}\hspace{1cm}(D)&lt;\frac{1}{n}$  $\frac{a^2_1+a^2_2+a^2_3+....+a^2_{n-1}}{a^2_n}=\frac{a^2_1}{a^2_n}+\frac{a^2_2}{a^2_n}+\frac{a^2_3}{a^2_n}+......+\frac{a^2_{n-1}}{a^2_n}&gt;(n-1)\frac{a^2_1}{a^2_n}\frac{a^2_2}{a^2_n}\frac{a^2_3}{a^2_n}......\frac{a^2_{n-1}}{a^2_n}$  and then no idea how to proceed.Any hints,please?
104983
$a _{1} = \frac{1}{2}, a _{2} = 1, a _{n}= \frac{1}{2} a_{n-1} + \sqrt{a _{n-2} }$  Show by the induction that it's $\le 4$ and also by the induction that $a_{n+1}-a_n \ge 0$  If it comes to first I don't know what to do and if it comes to second I tried many different ways but I couldn't prove it.
30495
Let $a_{n+1}=a_n+\frac{n}{a_n}$ and $a_1&gt;0$. Prove $\lim\limits_{n\to \infty} n(a_n-n)$ exists.  In my view, maybe we can use  $${a_{n + 1}} = {a_n} + \frac{n}{{{a_n}}} \Rightarrow {a_{n + 1}} - \left( {n + 1} \right) = \left( {{a_n} - n} \right)\left( {1 - \frac{1}{{{a_n}}}} \right).$$ And then  $${a_n} - n = \left( {{a_1} - 1} \right)\prod\limits_{k = 1}^{n - 1} {\left( {1 - \frac{1}{{{a_k}}}} \right)} .$$ By Stolz formula, we have  \begin{align*} &amp;\mathop {\lim }\limits_{n \to \infty } n\left( {{a_n} - n} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{\frac{1}{{{a_n} - n}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{\frac{1}{{{a_{n + 1}} - \left( {n + 1} \right)}} - \frac{1}{{{a_n} - n}}}}\\  = &amp;\mathop {\lim }\limits_{n \to \infty } \frac{{\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right]\left( {{a_n} - n} \right)}}{{{a_n} - {a_{n + 1}} + 1}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right]\left( {{a_n} - n} \right)}}{{ - \frac{n}{{{a_n}}} + 1}}\\  = &amp;\mathop {\lim }\limits_{n \to \infty } {a_n}\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right]. \end{align*} And how can we continue?
106366
In my book this is termed as a theorem and the proof given is as follows :-  $$\begin{align} \lim_{x \to a}\left(\frac{x^n - a^n}{x - a}\right) &amp;=\lim_{x \to a}\left(\frac{(x - a)*(x^{x-1} + x^{n-2}*a + x^{n-3}*a^2 + x^{n-4}*a^3 + \cdots + x^1*a^{n-2} + a^{n-1})}{x - a}\right) \\ &amp;=(a^{n-1} + a*a^{n-2} + \cdots + a^{n-1}) \\ &amp;=(a^{n-1} + a^{n-1} + \cdots + a^{n-1}) \\ &amp;=(n*a^{n-1}). \end{align}$$  Everything made sense to me except   $$x^n - a^n = (x - a)*(x^{x-1} + x^{n-2}*a + x^{n-3}*a^2 + x^{n-4}*a^3 + \cdots + x^1*a^{n-2} + a^{n-1})$$  Somebody please enlighten me on this topic.
70071
This is a reformulation of a deleted question:  If  $a_1 &gt; 0$ and $a_2 &gt; 0$ and $2a_{n+2} \le a_{n+1}+a_n$,  show that  $\lim \sup a_n \le \frac23 a_2 + \frac13 a_1$.  My proof involves showing that $a_{n+2} \le u_{n} a_{n+1} + (1-u_n)a_n $ where $u_n \to \frac23$, and I wondered if there is a simpler proof.  I will post my proof in a couple of days if no one does.
15864
A sequence $a_n$ is defined as $a_1=1$ and $a_{n+1}=\sqrt{a_1+a_2+\cdots+a_n}$ .Prove that $ \lim\limits_{n \rightarrow \infty} \frac{a_n}{n}=\frac{1}{2}$  I have no idea how to approach this. But I have a feeling that Cesaro's lemma may come in handy
14641
I would like to confirm a solution. The question goes as:  Show that $$\frac{\sqrt{3}}{\sin20^{\circ}}-\frac{1}{\cos20^{\circ}}=4$$  Firstly I combined the two terms to form something like: $$\dfrac{\sqrt{3}\cos20^{\circ} - \sin20^{\circ}}{\sin20^{\circ}\cos20^{\circ}}.$$  Clearly, the numerator is of the form $a\cos\theta+b\sin\theta$, and can thus be expressed as $R\sin(\theta+\alpha)$, where $R=\sqrt{a^2+b^2}$, and $\alpha=\arctan(\dfrac{a}{b})$. Following this method, I obtain something like this: $$\dfrac{2\sin(20^{\circ}-60^{\circ})}{\sin20^{\circ}\cos20^{\circ}},$$ which results in the expression to be equal to $-4$. But, we took $R$ to be the principal square root, which was $2$. If we take it to be the negative root, $-2$, it follows that the expression is equal to $4$. So, can this expression have two values?   Edit: I would also like to mention how I computed the numerator:  Numerator:   $\sqrt{3}\cos20^{\circ} - \sin20^{\circ}$.  Now, this can be represented as $R\sin(\theta+\alpha)$. Computation of $R$: $R=\sqrt{a^2+b^2}\implies R=\sqrt{(3+1)}=2$ Computation of $\alpha$: $\tan\alpha=\dfrac{a}{b}\implies \tan\alpha=\dfrac{\sqrt3}{-1}\implies \alpha=-60^{\circ}$, because $\tan(-x)=-\tan x$, and in this case $\tan\alpha=-\sqrt3$, and $-\tan(60^{\circ})=-\sqrt3$, so $\alpha=-60^{\circ}$.  Hence, we conclude that numerator:$$2\sin(20+(-60))=2\sin(-40)=-2\sin(40^{\circ})$$
21646
Prove that $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational   My attempt:-     Suppose $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is rational, then for some $x\in\mathbb{Q}$ we have $$\sqrt{6}-\sqrt{2}-\sqrt{3}=x$$ Rewriting this equation as $$\sqrt{6}-x=\sqrt{2}+\sqrt{3}$$ and now squaring this we get $$ 6-2x\sqrt{6}+x^2=5+2\sqrt{6}$$. This implies that $$\sqrt{6}=\frac{x^2-1}{2+2x}$$ but this is absurd as RHS of the above equation is rational but we know that $\sqrt6$ is irrational. Therefore , $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational. Does this look good? Have I written it properly? Is there any other proof besides this..like one using geometry? Thank you.
36314
Because $\sin'(x)=\cos(x)$ we can prove that $\arcsin'(x)=\frac{1}{\sqrt{1-x^2}}$. but, by definition we have $$\arcsin'(x)=\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}\tag{1}$$ therefore, $$\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}=\frac{1}{\sqrt{1-x^2}}\tag{2}$$ My question: Can we prove $(2)$ without using $(1)$? How?
72759
Suppose that $a_{1}&lt;a_{2}&lt;…&lt;a_{n}$. Prove that the equation  $ \frac{1}{x-a_{1}}+\frac{1}{x-a_{2}} +…+\frac{1}{x-a_{n}}=c$ has exactly $n-1$ roots if $c=0$ and $n$ roots if $c\neq 0$.    I am absolutely not sure but I think I need to use Rolle's theorem on this. Multiply both sides by $x-a_{1}$ so we get:  $1+\frac{x-a_{1}}{x-a_{2}}+…+\frac{x-a_{1}}{x-a_{n}}-c(x-a_{1})=0$. Now it is defined for $a_{1}$ but not for $a_{2},…,a_{n}.$. If we consider the interval $(a_{1}-\varepsilon,a_{n}+\varepsilon), \varepsilon&gt;0$ we get..nowhere. I want to create an interval with endpoints that are solutions to the equation. Yet I don't know how to prove that there are exactly $n$ roots(Fundamental theorem of algebra can't be used).