I have the following sequence $(x_{n})_{n\geq 1}$ ; $x_{n+1}=x_{n}^2-x_{n}+1$ I need to find $x_1$ such that  $(x_{n})_{n\geq 1}$ is convergent.  I found that $(x_n-1)^{2}&gt;0$ so $x_n$ is increasing.  Also, If I note the limit with $L$ I get $L^2-2l+1=0$ so $L=1$  If $x_1&gt;1$ and $x_n$ is increasing result that $x_n&gt;1$  How to continue?
121684
Suppose that $x_0 = \alpha \in \mathbb{R}$ and $x_{n+1} = x_n ^2-x_n +1$.  I am asked to study the boundedness of $(x_n)$ and then asked if $(x_n)$ converges. How can I show that $(x_n)$ is bounded?  I have noted that $$x_{n+1}-x_n = (x_n-1)^2\geq 0$$ so $x_n$ is increasing.  Suppose that $(x_n)$ converges, then $l = l^2-l+1$ so $l = 1$. This means that the sequence converges only when $\alpha \leq 1$. And so $(x_n)$ is bounded above by $1$ if $\alpha\leq 1$ (since an increasing sequence converges to its supremum) otherwise the sequence is not bounded above.  Is my reasoning correct?
114196
Let $x_n$ be the unique non-negative real solution of a equation $x^n+x^{n-1}+x-1=0$. Prove the sequence {$x_n$} is increasing and converges to 1.  I found out that $ 0 &lt; x_n &lt; 1$ but just that..
107521
I am assuming my teacher wants us to conclude they converge by being bounded and monotone, since part (i) is to show that $a_n$ is monotone increasing and $b_n$ is monotone decreasing.  I am confused though because I have never worked with a sequence defined with... another sequence.
98646
Show that the sequence $$\sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},...$$ converges and find its limit. I put the sequence in this form , $(x_n)$ where $$\large x_n=2^{\Large\sum_{k=1}^{n}\left(\frac{1}{2^k}\right)}$$ I want to use Monotonic Convergence Theorem to show. But I stuck at proving the sequence is bounded above by 2. I manage to prove the sequence is an increasing sequence. Anyone can guide me ? I try to use induction to prove but i stuck at inductive step.
41102
$a_n$ is a sequence that satisfies the following contractive condition;  $|a_{n+2}-a_{n+1}| \le c|a_{n+1}-a_n|, n\ge 1 $ for some $ 0 \lt c \lt 1.$ Show the sequence $a_n$ is a Cauchy sequence and converges.  Consider the sequence $x_n$ given by $x_1=2$ and $x_{n+1} = 2+ \frac{1}{x_n}$   Show that $x_n$ is bounded below by 2 Show that $x_n$ satisfies the contractive condition in the first part and converges Find lim $x_n$   Not looking for the answers. Just strong hints to guide me to the answer. The first part of the problem itself, makes sense. I'm unsure on the steps needed to prove this though. The second part for (1), would I want to use induction to show that it keeps getting closer to 2 but never reaches it. For part (2), I may be way off, but as simple as plugging into the contractive condition above? I am confused on part 3.  A lot here, I appreciate any sort of help. Thanks
112504
I've already proved that for any $p &gt; 0$ and for any $\alpha \in \mathbb{R}$, the sequence $\frac{n^\alpha}{(1 + p)^n}$ converges to $0$.  Now, I want to prove that $\lim_{n \to \infty} x^n = 0$ as long as $|x| &lt; 1$.  I've split it into two possible cases:  (1) $0 &lt; x &lt; 1$.  Let $p = \frac{1}{x} - 1$, so that $x = \frac{1}{1 + p}$ and $p &gt; 0$ (since $x &lt; 1$). Then we let $\alpha = 0$ to see that $$\lim_{n \to \infty} x^n = \lim_{n \to \infty} \frac{n^0}{(1 + p)^n} = 0$$  (2) $-1 &lt; x &lt; 0$.  Here we cannot use the sequence $\frac{n^\alpha}{(1 + p)^n}$ again, since if we let $p = \frac{1}{x} - 1$, then $p$ isn't necessarily positive (for example, if $x = - \frac{1}{2}$, then $p = -3 &lt; 0$).  How can I prove Case (2)?
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 $\sqrt{3}$?
52733
If $x\in \Bbb R$ ; then is the sequence $\{a_n\}$ where $a_1=x$; $a_{n+1}=\cos (a_n)$ convergent?  Obviously $|a_n|\le1$  and hence $(a_n)$ is bounded.  Also $f(x)=\cos x$ is decreasing for $x&gt;0$.  But here $x\in \Bbb R$ .How to proceed here?Please help.
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.
64951
Let $(a_n)_{n\in\mathbb{N}}$ be a sequence such that lim inf $|a_n|=0$. Prove that there is a subsequence $(a_{n_{k}})_{k\in\mathbb{N}}$ such that $\sum_{k=1}^\infty a_{n_k}$ converges.  I'm thought about showing that you can make a subsequence that is smaller then $1/n^2$. Is this the right strategy ? Any hints ? Should I use the Cauchy criterion ?
100616
$f_n(x) = x - x^n$ for $x\in [0,1]$. Does the sequence converge pointwise or uniformly on $[0,1]$?   We also have to prove why it is or isn't pointwise or uniformly convergent.   I have worked out that on $x\in[0,1)$, the limit as $n$ approaches infinity is $x$. And when $x=1$ the limit is $0$.   So this means its pointwise convergent right? But how do I prove this?  Also how on earth do I prove it is not uniformly convergent? (I'm just assuming it's not).
123238
Let $x_n$ be a sequence of real numbers.  Show that if $\sum_{n=1}^{\infty}x_n $ converges, then $\lim_{n\rightarrow\infty}\sum_{k=n}^{\infty}x_k=0.$  This is what I have done. We have by the Cauchy Criterion, for every $\epsilon&gt;0$ there is an $N\in\mathbb{N}$ such that $n\ge N \ \implies |\sum_{k=n}^{\infty}x_k|&lt;\epsilon.$ Taking the limit as $n \rightarrow \infty$ we get the desired result.  Are there any problems with my proof?
50518
Find all x for which $\sum_{n=1}^n x^{n^2} n!$ is convergent  So I tried using the ratio test and got $\lim_{n\to \infty} x^{2n+1}(n+1)$ but I don't know how to proceed from that.   I also tried using the root test which gave me $\lim_{n\to \infty} x^{n}(n!)^{1/n}$ which wasn't really helpful either...
93622
I am trying to prove some theorem, and in the proof of it I have to check the fact that if the sequence $\{a_n\}_{n\in \mathbb{N}}$ converges, then $$\lim_{n\to \infty} n(a_n - a_{n-1}) = 0. $$ I think that this is true because I have tried with many different sequences and it holds, but I do not know how to prove it.  If this is not true, what is a counterexample for it?
4506
Prove that $(a_{n})_{n \in \mathbb{N}}$ is convergent with $a_{0}=1$,   $a_{n+1}= \sqrt{2a_{n}}$ and calculate its limit.      You got the info that the square-root-function is strictly monotonic   increasing and also continuous.   I'm very confused how this could be solved because there isn't really a sequence given here. It seems to be a recursive one and I also don't know how to form $a_{n+1}$.  Could I form it to $a_{n}+ a_{1}$ ?(I don't care if it's useful here but I'd like to know if this would be correct at first.)    I have started by forming to $a_{n}$:  $$2a_{n} = (a_{n+1})^{2}$$  $$a_{n}=\frac{(a_{n+1})^{2}}{2}$$  Then I also formed to $a_{1}:$  $$a_{n}+ a_{1} = \sqrt{2a_{n}}$$  $$a_{1} = \sqrt{2a_{n}} - a_{n}$$  But it seems like this won't lead to anything.. Also I don't know what we can do with the given info, $a_{0} = 1$, the continuity and monotonic increasing...
64925
Let $\{a_n\}$ be a sequence of real number, let $\{n_k\}$ be an increasing sequence of positive integer and let $A_n=a_{n_k}+a_{n_{k+1}}+...+a_{n_{k+1}-1}$ for $k \in N$. Prove that if $\sum_{n=1}^\infty a_n$ converges then $\sum_{k=1}^\infty A_k$ converges. Is the converse true?  I know that $\sum_{n=1}^\infty a_n$ converges implies that $lim _{n_&gt;\infty} a_n =0$, can I conclude that $A_n$ ->$0$ as well, even if I can conclude that, it still doesn't prove that $\sum_{k=1}^\infty A_k$ converges, am I correct?
79337
Suppose $a_0$ is an arbitrary positive real number. Define the sequence $\{a_n\}$ by $$a_{n+1}=\frac{a_n+2}{a_n+1}$$ for all $n\geq0$. I have to prove that $\{a_n\}$ converges.  My attempt: If  $a=\lim_{n\to\infty}{a_n}$ exists, then it should be a solution to $$a=\frac{a+2}{a+1}$$ which is $\sqrt2$. Thus I need to show that $|\sqrt2 - a_n|$ gets arbitrarily small for large $n$. I tried to prove that $|\sqrt2-a_n|&lt;|\sqrt2-a_{n-1}|$ but couldn't.