Let $x\in(0,\pi/2)$ and $\{a_n\}_{n\in\mathbb N}$ defined recursively as follows: $$ a_0=x, \quad \text{and} \quad a_{n+1}=\sin(a_n). $$  Show that $$ \lim_{n\to\infty}{n\,a_n^2}=3. $$  Note. There is a very strong numerical evidence that the above is valid, and the limit is independent of the initial choice of $a_0$. I have no idea where to start from.
17172
Consider the sequence defined as  $x_1 = 1$  $x_{n+1} = \sin x_n$  I think I was able to show that the sequence $\sqrt{n} x_{n}$ converges to $\sqrt{3}$ by a tedious elementary method which I wasn't too happy about.  (I think I did this by showing that $\sqrt{\frac{3}{n+1}} &lt; x_{n} &lt; \sqrt{\frac{3}{n}}$, don't remember exactly)  This looks like it should be a standard problem.   Does anyone know a simple (and preferably elementary) proof for the fact that the sequence $\sqrt{n}x_{n}$ converges to $\sqrt{3}$?
75676
Proposition: Let $(a_n)_{n=1}^\infty$ be a sequence of real numbers. Show that $\limsup_{n\to\infty} (a_n)\leq \sup(a_n)_{n=1}^\infty$.   Any hints on how to prove it will be appreciated.
52522
Question: Prove or disprove that $\lim_{n \to \infty} {n \choose k} / a^n = 0$ for all $k \in \mathbb{Z_+}, a \in \mathbb{R}: a&gt; 1$.  Comment: When I was falling asleep last night I was thinking about increasing integer sequences like triangle numbers $(0,1,3,6,10,15,\cdots)$ and  $(0,1,4,10,20,35,\cdots)$ or in short any integer sequences of the type $\left({0 \choose k}, {1 \choose k}, {2 \choose k}, \cdots\right)$ for some integer $k$ and I was wondering if for some $k$ that there is exponential growth at the limit, hence my question. I suppose not, but I could not prove it directly.   What if the $k$ constant increases as a function of $n$, for example if $k = \lceil n/2 \rceil$ (giving the sequence $\left({0 \choose 0}, {1 \choose 1}, {2 \choose 1}, {3 \choose 2}, \cdots\right)$)?
4438
Prove that if $\lim_{n \to \infty}z_{n}=A$ then:  $$\lim_{n \to \infty}\frac{z_{1}+z_{2}+\cdots + z_{n}}{n}=A$$ I was thinking spliting it in: $$(z_{1}+z_{2}+\cdots+z_{N-1})+(z_{N}+z_{N+1}+\cdots+z_{n})$$ where $N$ is value of $n$ for which $|A-z_{n}|&lt;\epsilon$ then taking the limit of this sum devided by $n$ , and noting that the second sum is as close as you wish to $nA$ while the first is as close as you wish to $0$. Not sure if this helps....
28088
I have the following problem.   I have to find the limit of $b_n = \frac{a_n}{n}$, where $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=l$  My approach:  I express $a_n$ in terms of $b_n$, i.e. $a_n=nb_n$ and $a_{n+1}=(n+1)b_n$  We look at the difference: $a_{n+1}-a_n=(n+1)b_{n+1}-nb_n$  Assuming that $b_n$ converges to a real number m, we see that:  $l=(n+1)m-nm$, from where I conclude that $m=l$.  What I'm left with is proving that $b_n$ is convergent which I'm not sure how to do.  Thanks in advance!
6645
Given $a_1 \in \mathbb R$ . Consider the   sequence  $\{a_n\}$, which is recursively  defined by$$ a_{n+1} =\left\{\begin{array}{ccc}   \dfrac {a_n}{2} &amp;  \text{if $n$ is even,}\\  \dfrac {1 + a_n }{2} &amp;  \text{if $n$ is odd}.\end{array} \right. $$  Find the sub-sequential limit of this sequence.  I think  for  odd  $n$  the  limit  point will be   $l= \frac{1+l}{2}$ now  $2l = 1+l$  , here $ l = 1$  I don't  know the  limit point for0 even  $n$.  Please help me.
24947
When $\lim_{n\to\infty} x_n = a$, and $\lim_{n\to\infty} y_n = b$, find the limit, $$\lim_{n\to\infty} \frac{x_1 y_n + x_2 y_{n-1} + \cdots + x_n y_1}{n}.$$ Thank you for your help in advance.
34311
How can I show:      $$\lim_{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$$   I've tried multiplying with its "conjugate" but that doesn't seem to help that much.  Thanks!
49258
The limit is   $$\lim_{x\rightarrow0} \frac{x-\sin_n(x)}{x^3},$$  where $\sin_n(x)$ is the $\sin(x)$ function composed with itself $n$ times:  $$\sin_n(x) = \sin(\sin(\dots \sin(x)))$$  For $n=1$ the limit is $\frac{1}{6}$, $n=2$, the limit is $\frac{1}{3}$ and so on.  Can we define a recurrent relation upon that given hypothesis? Also, how do I involve $n$ into calculation, because the final limit will depend on it?  Any suggestions on how to tackle this?  Thank you!
75948
How can I find the limit of the following recursive formula?  $a, b \in \mathbb{R}$ $a_0=a; a_1=b; a_n= \frac{1}{2}(a_{n-1}+a_{n-2})$ $\lim\limits_{n \rightarrow \infty}{a_n} = ?$  Many thanks in advance.
15740
Ratio and root tests won't help. And I can't use the comparison test because $ |a_n| $ is not necessarily smaller than $ n^2a_n $.   Can I use limits? We know:    $\lim\limits_{n \to \infty} a_n = 0 $ $\lim\limits_{n \to \infty} |a_n| \ne 0 $   And we need to prove:    $\lim\limits_{n \to \infty} n^2a_n \ne 0 $   Any ideas/hints?
88110
I am given the problem: Let $a_n&gt;0$, prove $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty \sin(a_n)$ converges.  We have this problem in a homework, and I don't believe it can be true. The statement being biconditional implies that when $\sum_{n=1}^{\infty} \sin(a_n)$ converges, so too does $\sum_{n=1}^\infty a_n$, but if we take $a_n=\pi n$ or even $a_n=\pi$, then $\sum_{n=1}^\infty \sin(a_n)$ converges while $\sum_{n=1}^\infty a_n$ diverges. Is my thinking incorrect?
91070
I'm struggling to find the limit at infinity of : $$\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n}), n\in\Bbb N$$ I know it is $1$ but I don't understand why this is wrong :  $\sin^2(\pi\sqrt{n^2 + n}) = \sin^2(\pi*n\sqrt{1 + \frac{1}{n}})$.  So $\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n}) = \lim \limits_{n\rightarrow\infty}\sin^2(\pi*n\sqrt{1 + \frac{1}{n}}) =  \lim \limits_{n\rightarrow\infty} \sin^2(\pi*n) = 0$.  Thanks.
120543
Show:       $$ \lim_{n\to \infty} \int_{0}^{\frac{\pi }{2}} \sin^n(x) = 0. $$   I showed this using the . I want other methods to prove it. Thank you.
105434
Good morning, i have a big problem with this proof. I haven't idea about this proof.   Problem:   Suppose $\lim_{n\rightarrow\infty}a_n =L  $ then $\lim_{n\rightarrow\infty}\ \frac{a_1+a_2+\cdots+a_n}n=L  $  I've tried this:  $| a_n-L|&lt;\varepsilon\Rightarrow L-\varepsilon&lt;a_n&lt;L+\varepsilon$ Please help!!
91488
Let $a\geq 0$ and $ b\geq 0$. Prove that $\lim_{n \to \infty} \sqrt[n]{a^n+b^n}=\max \{a,b\}$. [Hint: Use the identity $(a^n -b^n)=(a-b)(\sum_{i=0}^{n-1}a^ib^{n-1-i})$]    I need some help! I cannot do it even with the hint... :(
81275
Question:  Let ${a_n}$ be a sequence in $\mathbb{R}$ with $\lim_{n\to\infty}a_n=a$  Prove that  $\lim_{n\to\infty}a_n^2=a^2$  Attempt: Without using any properties of limits (this is a question in a section of the book that came before any list of any limit theorems)  By definition:  $|a_n^2-a^2|&lt;\epsilon$   $|a_n^2-a^2|$ $\leq$ $|(a_n+a)(a_n-a)|$ $\leq$ $|(a_n+a)| |(a_n-a)|$ $\leq$   $(|a_n | + |a|)(|a_n -a|)$ $&lt;$ $\epsilon$  And since I know that $\lim_{n\to\infty}a_n=a$, I'll subsitute that in for $|a_n|^2$ as $a^2$   $\therefore$ $a^2-|a|^2 &lt; \epsilon$ which is true since $a^2-|a|^2$ in my case.  I wonder if this is a possible solution?