Let $a_1,a_2,..,a_n$ be sequence of real numbers such that $a_{n+1}=a_{n}+\sqrt{1+a_n^2}$ and $a_0=0$.  How to evaluate $\lim_{ n\to \infty  }\frac{a_n}{2^{n-1}}$ ?
12783
Be $(x_n)_n\ge 1 $  such that $x_1=1$ and $x_{n+1}=x_n+\sqrt{x_n^2+1}$ for every $n\ge 1$. Prove that the sequence $y_n=(2^n/x_n)_n \ge 1$ is convergent and find it s limit.  Being positive and decreasing, $y_n$ is clearly convergent. But finding it's limit really put me intro trouble. Any help?  Thank you!
116751
Assume $\{a_n\}$is a sequence which  satisfies the following recursion formula: $$a_{n+1}=a_n+\frac{2a_{n-1}}{n+1}~(n\geqslant 1).$$ and $a_0=\pi,a_1=\pi^2 $. How to compute $\displaystyle\lim_{n\to \infty}\frac{a_n}{n^2}$?
87031
Let $(x_n)_n$ be a sequence of real numbers such that $$\lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0.$$ Show that $(x_n)_n$ must also converge to zero.     $\textbf{My Attempt:}$   My original idea was to use the triangle inequality and the fact that $$\lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0$$   to show that the sequence $(x_n)_n$ is Cauchy and hence converges to some real number $a$, which would give me $$\lim\limits_{n\rightarrow \infty}x_n=a=\lim\limits_{n\rightarrow \infty}x_{n+1}\implies a-\dfrac a 2= \lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0\implies a=0.$$ However, I'm having trouble showing that $(x_n)_n$ is Cauchy.   Any suggestions/hints would be greatly appreciated. Thanks!
63450
Suppose $|a|&lt;1$ be a real number. How to solve the following limit? $$\lim\limits_{n\to \infty}(1-a)(1-a^2)\cdots(1-a^n).$$
69669
For $a,b,c&gt;0$, Find $$ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$$  how can I find the limit of sequence above?  Provide me a hint or full solution. thanks ^^
108876
Let a sequence be $a_1=1;a_{n+1}=(n+1)(1+a_n)$ If $P_n=\prod_1^n(1+a_i^{-1})$ then $$\lim_{n\to\infty}P_n $$is ?    I did: $$P_n=\prod_1^n\frac{(1+a_i)}{a_i} =\frac{a_{n+1}}{a_1}\prod1^{n-1}\frac{(1+a_i)}{a_{i+1}} =\frac{a_{n+1}}{a_1}\prod_1^{n-1}\frac1{i+1}=\frac{a_{n+1}}{n!}$$ Now how do I find the limit?
21647
I need some suggestions, hints for the limit when $a \to 1^{-}$ of $$\sqrt{\,1 - a\,}\,\sum_{n = 0}^{\infty}a^{n^{2}}.$$
55445
Let $\{a_n\}$ be defined by $a_1 =1  $ and $a_{n+1} = 1 + {\dfrac{1}{a_n}}$ with $n \in N$. Show that $a_{n+1}= 1 + {\dfrac{1}{a_n}}$ converges.  I know the limit but how can I show that is a Cauchy sequence or that this sequence converges.
99697
I want to prove that the sequence defined by $\{x_1=1,\,x_{n+1}=\frac {x_n}2+\frac 1{x_n} \}$ has a limit.  By evaluating the sequence I notice that the sequence is strictly monotonically decreasing starting from $x_2=1.5$.  It seems to suggest itself to prove that the sequenced is bounded by $1\le\big( x_n \big)_{n\ge1} \le 1.5$ and to prove that it is strictly monotonically decreasing starting at $x_2$ which would imply convergence.  How would I proceed and could one prove the existence of the limit without first evaluating the values of the sequence to see how the sequence behaves?
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?
24384
Define the recursive sequence $(a_n)$ as $a_{n+1} = a_n^2 + a_n$ where $n \in \mathbb{N}$. We want to find $a_1 \in \mathbb{R}$ such that $(a_n)$ converges to a limit that depends on $a_1$.   Attempt: Let $a_1 = k$, then $a_2 = k^2 + k$, $a_3 = (k^2 + k)^2 + (k^2 + k)$, ... clearly if we have |k| > 1 then the sequence $(a_n)$ will diverge. From this we deduce the interval of convergence must be when $|k| \le 1$. Suppose $(a_n) \to L$ then $a_{n+1} = a_n^2 + a_n$ $\implies L = L^2 + L \implies L = 0$. Surely if $k = 0$ this will occur as $a_n = 0$ for all n. But it is not clear to me if there are other $k$ such that the recursive sequence converges.
103599
can anyone help me with this, I got stuck on it.  let $a_n$ and $b_n$ (for $n$ a non-negative integer) be two sequences of real numbers such that $\lim_{n\to \infty}a_n=a$ and $\lim_{n\to \infty}b_n=b$. prove that $\lim_{n\to \infty} \frac{a_0b_n+a_1b_{n-1}+....+a_nb_0}{n}=ab$.
727
Setting $a_n=\sqrt{2+\sqrt{2+\sqrt{2+}}}\ $ I get that $$a_{n+1}=\sqrt{2+a_n} \quad \quad a_1=\sqrt{2}.$$ Clearly all numbers in the sequence are positive and we see that $a_n&lt;a_{n+1} \ \forall \ n$ which implies that the sequence is strictly increasing. To show that it's convergent, I need to show that it's bounded above.  We can use the help-function $f(x)=\sqrt{2+x}$ such that $a_{n+1}=f(a_n).$ But since $$f'(x)=\frac{1}{2\sqrt{2+x}}=0\Leftrightarrow\text{no real solutions,}$$  $f(x)$ never flattens out or decreases, so it can't be convergent?
108652
Solving some problems regarding limits and sequence convergence, i stumbled upon a task, and it's solution relies on, and i quote: "We now use a well-known theorem : $$\lim_{n \to +\infty }{\sqrt[n]{a_1 a_2 \ldots a_n}} = \lim_{n \to +\infty}{a_n}$$  This isn't really intuitive (at least to me) and I don't know how to prove it. The original task was to find the limit of  $$\lim_{n \to +\infty }{\sqrt[n]{\bigg{(}1+\frac{1}{1}\bigg{)} \bigg{(}1+\frac{1}{2}\bigg{)}^2 \ldots \bigg{(}1+\frac{1}{n}\bigg{)}^n}} $$ which of course, using the expression above is just $e$.
30839
Let $a:\mathbb{N}\to\mathbb{R}^+$ be a sequence such that $a_{m+n}\leq a_ma_n$. Prove that $\sqrt[n]a_n$ converges and $$\lim_{n\to\infty}\sqrt[n]a_n=\inf_{n\in\mathbb{N}}\sqrt[n]a_n.$$  So far all I have is $\sqrt[n]a_n\leq a_1$ (from $a_{m+n}\leq a_ma_n$ by induction). If it is true that $\sqrt[n]a_n$ monotone decreasing, we are done but I couldn't show it. Should I do something with $\liminf$ and $\limsup$?
115529
I found the following question in a book:-  $Q:$Let $a_1, a_2, ... , a_n$ be a sequence of real numbers with $a_{n+1}=a_n+\sqrt{1+a_n^2}$ and $a_0=0$. Prove that $$\lim_{n \to \infty} \frac {a_n}{2^{n-1}}=\frac 4{\pi}$$ I tried many things none of which seemed fruitful. First thing I did was to define $a_n=2^{n-1}b_n$. Substituting this into the condition, we get $$2^{n}b_n=2^{n-1}b_n+\sqrt{1+2^{2n-2}b_n^2}\implies b_{n+1}=\frac {b_n}2+\sqrt{\frac1{2^{2n}}+\left(\frac {b_n}2\right)^2}$$ Simplifying this gives $$b_{n+1}^2-b_{n+1}b_n=\frac1{2^{2n}}\implies b_{n+1}(b_{n+1}-b_n)=\frac1{2^{2n}}$$ This doesn't lead anywhere. One lead that I got was by substituting $a_n=\tan \theta_n$. This gives $$\begin{align}a_{n+1}&amp;=\tan \theta_n+\sec \theta_n\\&amp;=\frac{\sin\theta_n+1}{\cos\theta_n}\\&amp;=\frac{1+\tan\frac{\theta_n}2}{1-\tan\frac{\theta_n}2}\\&amp;=\tan\left(\frac {\theta_n}2+\frac{\pi}4\right)\end{align}$$ Hence $$\tan\theta_{n+1}=\tan\left(\frac {\theta_n}2+\frac{\pi}4\right)\implies \theta_{n+1}=\frac {\theta_n}2+\frac{\pi}4\implies \theta_n=\frac {\pi}2+c\left(\frac 12\right)^n$$From initial conditions we get $\theta_0=0\implies c=-\frac{\pi}2$. Therefore $$\lim_{n \to \infty} \frac{a_n}{2^{n-1}}=\lim_{n\to\infty}\frac{\tan\left(\frac {\pi}2-\frac{\pi}2 \left( \frac 12 \right)^n\right)}{2^{n-1}}=\lim_{n\to\infty}\frac{2\left(\frac12\right)^n}{\tan\left(\frac {\pi}2\left(\frac12\right)^n\right)}=\lim_{n\to\infty}\frac{2\left(\frac12\right)^n}{\frac{\pi}2\left(\frac12\right)^n}=\frac4{\pi}$$ Is this proof valid and are there any other ways of doing it?
91702
Define a sequence of real numbers recursively as follows. Let $a_1 = 1$ and $a_{n+1} = 1 + \frac{1}{1+a_n}$.  First, show the sequence is not monotonic.  Second, show that $a_n \geq 1$ for all $n$ and conclude that the sequence is super Cauchy.