A sequence $\{x_n\}$ is given by $x_n= a_n-\sqrt{2n}$ and $a_n$ is given by $a_1=1$ and $a_n=a_{n-1}+\frac{1}{a_{n-1}}$.  Please help. I could not find a way out.
18261
Today, we had a math class, where we had to show, that $a_{100} &gt; 14$ for  $$a_0 = 1;\qquad a_{n+1} = a_n + a_n^{-1}$$  Apart from this task, I asked myself: Is there a closed form for this sequence? Since I didn't find an answer by myself, can somebody tell me, whether such a closed form exists, and if yes what it is?
78041
I would like some advice on how to solve problems like the following: Let $(x_n)$ be a sequence defined by $x_1= 3$ and $x_{n+1} = \frac{1}{4-x_n}$. Prove that the sequence converges.  My strategy is to use the Monotone Convergence Theorem, but I am having trouble showing that the sequence is decreasing and bounded below. Here's my work so far:  Decreasing: The first 4 values are $3,1,1/3,3/11$, so let's assume $x_n \leq x_{n-1} \leq \ldots \leq x_1$. Want to show $x_{n+1} \leq x_n$. We have $x_{n+1} = \frac{1}{4-x_n}$. I want to have an upper bound for the RHS, but can't find one and don't really know where to go from here.  Bounded below: I wanted to show that all values are positive, but if we assume $x_n &gt; 0$, that doesn't rule out $x_{n+1}$ from being negative.
6939
We define  $$a_{n+1}=\frac{1}{1+a_n}, a_0=c&gt; 0$$   I stumbled across that sequence and Mathematica gives me that it converges against $\frac{1}{2} \left(\sqrt{5}-1\right)$ which doesn't even depend on $a_0$. Normally I show that sequences converge by seeing that they are monotonic and then the limit can be easily found by setting $a_n=a_{n+1}$, but this one seems to be alternating. Also by checking OEIS I noticed that $a_n=1/g(n)$ where $g(n)$ gives the $n+1$'th .  I would be glad if someone can show me how to show that such sequences converge and how to find the limit, also maybe this is a well known sequence, as it has a quite simple form, too bad that google is very bad for looking for sequences and OEIS doesn't mention anything.
545
Let $s_n$ be a sequence defined as given below for $n \geq 1$. Then find out $\lim\limits_{n \to \infty} s_n$.   \begin{align}   s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx  \end{align}  I have written a solution of my own, but I would like to know it is completely correct, and also I would like if people post more alternative solutions.
64743
Let $(a_n)_n$ be a convergent sequence of integers , what can we say about $(a_n)_n$?   (I don't understand what is meant by this question)
103852
Let $\{f_n(x)\}_{n\in \Bbb N}$ be a sequence of polynomials defined inductively as   $$\begin{matrix} f_1(x) &amp; = &amp; (x-2)^2 &amp; \\ f_{n+1}(x)&amp; = &amp; (f_n(x)-2)^2, &amp;\text{ for all }n\ge 1.\end{matrix}$$      Let $a_n$ and $b_n$ respectively denote the constant term and the coefficient of $x$ in $f_n(x)$. Then         $(\text A)\,\, a_n=4, b_n=-4^n$   $(\text B)\,\,a_n=4, b_n=-4n^2$   $(\text C)\,\, a_n=4^{(n-1)!}, b_n=-4^n$   $(\text C)\,\, a_n=4^{(n-1)!}, b_n=-4n^2$      Please help me how to solve this problem. I am not able to solve it.  The problem can be found on this .
99792
Let $a\ge1$ and define the sequence $(x_n)$ recursively by: $$x_1 = \sqrt{a}$$ $$x_{n+1}= \sqrt{a+x_n}$$  Here's what I did:  Plugging in some values makes it seem as if the sequence is increasing. I first solve the equation $t=\sqrt{a+t}$, equivalently $t^2 -t -a =0$, to get an idea what's going on. The roots are $$t_{1,2}= \frac{1 \pm \sqrt{1+4a} }{2}.$$ Therefore, between the roots, that is for $$t\in \left( \frac{1 - \sqrt{1+4a} }{2}, \frac{1 + \sqrt{1+4a} }{2}\right)$$  we have $t^2-t-a \le 0 \iff t \le \sqrt{a+t}$. Here's my attempt at showing inductively that $$\forall n\in \mathbb{N}, \ x_n \le \frac{1 + \sqrt{1+4a} }{2}$$  For $n=1$, the statement is true (simple arithmetic). Suppose it's true for $n$. Consider  $$x_{n+1}= \sqrt{a+x_n}\le \sqrt{a+\frac{1 + \sqrt{1+4a} }{2}}=\frac{1 + \sqrt{1+4a} }{2},$$  because $\frac{1 + \sqrt{1+4a} }{2}$ satisfies $x=\sqrt{x+a}$.  So this establishes the sequence $(x_n)$ is bounded. Now, by induction again, I try to show it's monotonically increasing. The base case $x_1\le x_2$ is obvious. Assume (strong induction) $x_{n-1}\le x_n$ for all $n&gt;1$. But since we have $$x_1\le x_n \le \frac{1 + \sqrt{1+4a} }{2}, $$   then $x_n$ lies in the appropriate interval, so $x_n \le \sqrt{a+x_n}=x_{n+1}.$  Is this proof correct? Can the condition on $a$ be relaxed to $a&gt;0$? Maybe even small negative $a$? And could someone provide me with another proof of convergence of $x_n$? I absolutely hate this one!
111703
If the sequence $\{a_n\}$ with $n∈\mathbb{N}$ is a Cauchy sequence,  then $\{a_n^2\}_{n∈\mathbb{N}}$ is also a Cauchy sequence. How do we prove 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.
38845
Does the recursive sequence $a_1 = 1, a_n = a_{n-1}+\frac{1}{a_{n-1}}$ converge?  Since the function $x+1/x$ is strictly monotonic increasing for all $x&gt;1$, I don't think that the limit converges, but I'm not sure. Can anybody tell me whether the sequence is converging or not?
22696
I'm stuck with the prove of the following recurrence sequence which was part of and old exam.  $a_1:=\frac{1}{2}, a_{n+1}:=a_n(2-a_n)$ for $n \in \mathbb{N}$   I have to show that $0 &lt;a_n &lt; 1$ and as a second part (b), that its monotonic increasing (That's where I got stuck) and as a thid part (c) the convergence and limit of the sequence  a) My approach was to use induction to show that the sequence is upper and lower bounded: Base case: $ n=1, a_1 &lt; 1 = \frac{1}{2} &lt;1 $ ok Inductive step: $n\rightarrow n+1$ $a_{n+2} = a_{n+1}(2-a_{n+1}) &lt; 1$      Then my idea was to rewrite it as using the given definition from $a_{n+1}$: $a_{n+2}=a_n(2-a_n)(2-(a_n(2-a_n))$  and I know that $a_n&lt;1$ Then I used the base case that $a_n=\frac{1}{2}&lt;1$ so that I've got $ 0 &lt; \frac{3}{4}*(2-\frac{3}{4}) = 0.9375 &lt;1$   Question: Is this right?  b) For the second part of the task, to show montonic increasing I know that $a_{n+1} \leq a_n$ or $a_{n+1}-a_n\leq 0$ $a_{n+1}=a_n(2-a_n)  \geq 0 \Leftrightarrow a_n-a_n^2 \geq 0 \Rightarrow a_n^2-a_n \geq 0$ Here I got stuck:    Question: How do I make the final conclusion?  c) Limes: $a_n \rightarrow a$ $a=a(2-a) \Leftrightarrow a=2a-a^2 \Rightarrow a^2-a=0$ Solving gives $a_1=1$ and $a_2=0$ Because the sequence is monotonic increasing it converges and it must be $&gt;0$ so the $\lim(a_n) =1$  Question: Is that conclusion "exact", I mean can I use this conlcusion or is there something missing?  Best Regards, Christoph
104218
I am triyng to solve this interesting problem, but I have no idea. I tried induction, but it didn't work. Let $(x_n)_{n \ge 1}$ be a sequence, where $x_1=1$ and defined as follows: $$x_{n+1}=x_n+ \frac {1} {3x_n^2},$$ for every $n \ge 1$. Show that there exists a real number $c \gt 0$ so as $x_n \lt \sqrt[3] {n+c}.$
23431
Is the following true? Let $x_n$ be a sequence with the following property: Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.  I guess that it is true but I am not sure how to prove this.
112564
$|a_{n+1}-a_n| \leq \frac{1}{2} |a_{n}-a_{n-1}|$. I'm trying to show that $(a_n)$ is convergent by showing that $(a_n)$ is Cauchy.  let $|a_2 - a_1 |  =x$.   $|a_3-a_2| \leq \frac{1}{2}x$  $|a_4-a_3| \leq \frac{1}{2}|a_3-a_2|\leq \frac{1}{4}x$  In general, for every $n$:  $|a_{n+1}-a_n| \le \frac {1}{2^{n-1}}x$  Let $\epsilon &gt;0$  for every $m,n &gt; 0 $, say $m&gt;n$ the following holds:  $$|a_m-a_n| = |a_m-a_{m-1} + a_{m-1} - a_{m-2} \dots a_{n+1}- a_n| \le$$  $$|a_m-a_{m-1}| + |a_{m-1} - a_{m-2}| + \dots |a_{n+1}-a_n| \leq \frac{x}{2^{m-2}} + \frac{x}{2^{m-3}} + \dots + \frac{x }{2^{n-1}}$$.  I can say that for big enough $n,m$, $|a_m-a_n|&lt;\epsilon$ and therefore, this sequence is Cauchy? It sounds not good enough to me yet I don't know what else to do here.
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 $\sqrt{3}$?
10678
I am preparing for Putnam and working through practice-like questions and I am boggled. I've got the sequence defined by $a_1 = 1$ and $a_{n+1} = 2a_n+\sqrt{3a_n^2-2}$ for any $n\in\mathbb{N}$ and I'm trying to show that every term is a positive integer. I am getting as far as  $a_{n+1}^2 -4a_{n+1}a_n + a_n^2 = -2$ but I can't figure out how to use the recursive relationship to show that all terms are integers. I've tried expanding the sequence in terms of $a_n$ or $a_{n+2}$ but it doesn't seem to get me far. Any hints or help would be greatly appreciated.
96657
What I've done so far: I have proved that this sequence is bounded below by 0, which is a very rough estimate. I know that the infimum is $\sqrt2$.  Anyway, the question first asks me to prove that the sequence decreases monotonically. And I've tried the following:  Suppose $a_{n+1} \le a_n$, then we would get $\frac{1}{a_n} \le \frac{a_n}{2}$. After some algebra, I ended up with $\frac{1}{a_n^2 +2}\le\frac{a_n^2 +2}{2}$. Now, $a_{n+2} - a_{n+1}=\frac{2a_n}{a_n^2 +2}-\frac{a_n^2 +2}{4a_n}$. I was looking for some similarities between these two expressions but it doesn't seem to work out.  A more general question is, if the sequence is defined recursively, what are some common strategies to find the limit? I just started studying analysis and this kind of questions kind of troubles me.  Thank you guys.