Let $\left (a_n  \right )_{n=1}^{\infty}$ be an infinite sequence.  $\left (a_n  \right )_{n=1}^{\infty}$ is defined by:   $a_1 = 0$   $\; \; \; a_{n+1} = \frac{1}{1+ a_n}\; \; \forall n \in \mathbb{N}$  I need to show that $\left (a_n  \right )_{n=1}^{\infty}$ converges and find it's limit. I want to do it by finding a formula for both $a_{2k}$ and $a_{2k-1}$ and then use some basic limit arithmetics rules, however I didn't manage to find a formula for both sub-sequences.  I would love to get some recommendations on how to find such formula and how to approach these kind of questions.   Thanks!
122332
I have had big problems finding the limit of the sequence $x_1=0 , x_{n+1}=\frac{1}{1+x_n}$. So far I've only succeeded in proving that for $n\geq2$: $x_n&gt;0\Rightarrow x_{n+1}&gt;0$  (Hopefully that much is correct: It is true for $n=2$, and for $x_{n+1}&gt;0$ exactly when $\frac{1}{1+x_n}&gt;0$, which leads to the inequality $x_n&gt;-1$ which is true by the induction assumption that $x_n&gt;0$.)  On everything else I failed to come up with answers that make sense (such as proving that $x_{n+1}&gt;x_n \forall n\geq1$). I'm new to recursive sequences, so it all seems like a world behind the mirror right now. I'd appreciate any help, thanks!
87800
I know what solution is $e^a$ but I don't know how to calculate this limit:  $$\lim \limits_{n \to \infty}\left(1 + \frac{a}{n}\right)^n=e^a$$  Can someone explain the steps to me?
103850
I've looked up several related questions, but they do not answer what I am looking for.  Please give link if this is a duplicate.  What I eventually want to know is why  $$\lim_{n\to\infty}\left(1+\frac1n\right)^n$$  converges to a finite value.  I mean, we have to know that the limit converges before we can call it a constant, right?  I know that $f(n)=\left(1+\dfrac1n\right)^n$ is strictly increasing, but how do we know it is bounded above?  Also, I tried using the binomial theorem to write  $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=\lim_{n\to\infty}\left[\sum_{k=0}^n\frac{n!}{n^k k! (n-k)!}\right],\tag{1}$$  but I am unsure how to prove that the sum converges for $n\to\infty,$ because the general term depends on both $n$ and $k.$    Here is what I think will work.  Let $a_{n,k}=\dfrac{n!}{n^k k! (n-k)!}.$  Then $a_{n+1,k+1}=\dfrac{(n+1)!}{(n+1)^{k+1} (k+1)! (n-k)!}.$  Using the Ratio Test, we get  $$\lim_{n\to\infty\atop k\to\infty}\left|\frac{a_{n+1,k+1}}{a_{n,k}}\right|=\lim_{n\to\infty\atop k\to\infty}\left[\frac{1}{k+1}\left(\frac{n}{n+1}\right)^k\right].$$  where the limit on the RHS is less than 1.    That may take care of the case where $n\in\mathbb{N},$ so I suppose that (1) can be generalized using "."
7390
Let $\{x_{n}\}_{n \in \mathbb{N}}$ and $\{y_{n}\}_{n \in \mathbb{N}}$ be two convergent sequences.  I am trying to show that  \begin{align} \lim_{n \rightarrow \infty} \frac{1}{n+1}\sum_{k = 0}^{n} x_{k}y_{n - k} &amp;= \left(\lim_{n \rightarrow \infty} x_{n}\right)\left(\lim_{n \rightarrow \infty} y_{n}\right). \end{align}  I think I just need some fancy application of summation by parts but I haven't quite figured it out yet.
8956
a) Let $a&gt;0$ and the sequence $x_n$ fulfills $x_1&gt;0$ and $x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})$ for $n \in \mathbb N$. Show that $x_n \rightarrow \sqrt a$ when $n\rightarrow \infty$. I have done it in two ways, but I guess I'm not allowed to use the first one and the second one is incomplete. Can someone please help me?  We already know $x_n \rightarrow \sqrt a$, so we do another step of the iteration and see that $x_{n+1} = \sqrt a$.  Using limit, $x_n \rightarrow x, x_{n+1} \rightarrow x$ (this is the part I think it's incomplete, don't I have to show $x_{n+1} \rightarrow x$, how?), we have that   $$x = \frac x 2 (1 + \frac a {x^2}) \Rightarrow 1 = a/x^2 \Rightarrow x = \sqrt a$$ b) Let the sequence $x_n$ be defined as $x_{n+1}= 1 + \frac 1 {x_n} (n \in \mathbb N), x_1=1$. Show that it converges and calculate its limit. &quot;Tip: Show that sequences $x_{2n}$ and $x_{2n+1}$ monotone convergent to the limit.&quot; I didn't understand the tip, how can this help me? Does it make a difference if the number is odd or even? Thanks in advance!
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?
38006
Assume that $\{a_n\}$ is a convergent sequence. How to use the definition of a limit of a sequence to prove that $\lim\limits_{n \to\infty}a_{n+1} = \lim\limits_{n\to\infty}a_n$?
46873
I am looking for nice ways of proving the divergence of the sequence $\left\{x_n\right\}_{n=1}^{\infty}$ defined by $$x_n=\sin{(n)}.$$ One (not so nice) way is to construct two subsequences: one where the indexes are picked such that they lie in the intervals  $$I_k=\left(\dfrac{\pi}{6}+2\pi(k-1),\dfrac{5\pi}{6}+2\pi(k-1)\right)$$ and one where they lie in $$J_k=\left(\dfrac{7\pi}{6}+2\pi(k-1),\dfrac{11\pi}{6}+2\pi(k-1)\right).$$ If ${x_n}$ converges, then all its subsequences must converge to the same limit, but here the first subsequence has all its values in the interval $\left[ \frac{1}{2}, 1\right]$ while the second has its values in $\left[-1,-\frac{1}{2}\right]$. Contradiction.  Filling the details of this proof is rather tedious and not very elegant. Anyone has a better idea?
61557
A problem I saw recently is to compute $$\lim_{n\to\infty} n^2 \left(\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\right)$$   I thought it would be fun to give this a go but it completely stumped me. I am not aware of any of the standard techniques used to compute $\infty\times (e-e)$ - type limits, bar possibly L'Hôpital's Rule. But I doubt that would be a good idea here, at best it would be ugly. I would appreciate solutions using any methods, the more elegant the better of course. Though I welcome any relevant hints as well if full solutions are discouraged.
50833
Let $\{a_n\}, {n\geq 1}$, be a sequence of real numbers satisfying $|a_n|\leq 1$ for all $n$. Define  $$A_n = \frac{1}{n}(a_1 + a_2 + \cdots + a_n),$$ for $n\geq 1$. Then find $\displaystyle\lim_{n \rightarrow \infty}\sqrt{n}(A_{n+1} − A_n)$ .    I proceed in this way  $$\lim_{n \rightarrow \infty}\sqrt{n}(A_{n+1} − A_n)=\lim_{n \rightarrow \infty}\sqrt{n}\left[\frac{1}{n+1}(a_1 + a_2 + \cdots + a_n+a_{n+1})-\frac{1}{n}(a_1 + a_2 + \cdots + a_{n})\right]=\lim_{n \rightarrow \infty}\left[{(na_{n+1}-a_1 - a_2 - \cdots - a_n})\frac{1}{\sqrt{n}(n+1)}\right]$$  Please help me to complete from here
15658
Given $0 &lt; a_1 &lt; \frac{\pi}{2}$, for $n\geq 1$ we define $$a_{n+1} = \dfrac{{a_n}}{\sin({a_n})}.$$ I need to show that $\lim_{n\rightarrow\infty}{a_n} = L$, and then I need to find $L$.  I can't find a way to bound $\{a_n\}_n$ so that  $$0 &lt;a_n &lt; \frac{\pi}{2}$$ I've tried to use every convergence test I know, but I always have the same problem which I want to bound $\{a_n\}_n$ and can't to that without bounding $\{\sin(a_n)\}_n$.
82879
I have to find $$\lim_{n\to\infty}{(\sqrt[3]{n+1}-\sqrt[3]{n})}$$ here what I did $$\lim_{n\to\infty}{(\sqrt[3]{n+1}-\sqrt[3]{n})}$$=$$\lim_{n\to\infty}{\left(\sqrt[3]{n\left(1+\frac{1}{n}\right)}-\sqrt[3]{n}\right)}$$ How can I proceed?
16469
I have seen a proof that shows $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n = e^x$$ by looking at the Taylor series expansion of $\ln(1+x)$ at $x=0$. To prove a theorem, my textbook uses the fact $$\lim_{n \to \infty} \left(1+\frac{a_n}{n}\right)^n = e^a$$ when $a_n \to a$. How can I prove this?
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.
107026
$\lim_{x\to\infty} a_n$ defined as $a_{n+1}=a_n+\frac{1}{a_n}$ with $a_1=1$.  I've tried writing out the first few terms and this sum is increasing, but progressively less-so over time. I'm just not certain how to go about caclulating the limit here. I feel that it does have a finite limit, but I'm not certain of what it is.
32999
Let $a_n$ be the sequence in $\mathbb{R}$ given by:  $$ a_1 = \sqrt{2}, \; \; a_{n+1} =\sqrt{2+ \sqrt{a_n} }. $$  Show that this sequence converges and find the limit.  I have already proven that this sequence is increasing and bounded by $2$, so the limit exists. Now, when I'm trying to find the limit, say  $L$, I get that $L$ satisfies the equation: $$L^2 -2 - \sqrt{L}=0, $$ but I don't actually know how to solve this equation, so I would like to know if there is an  easier way to find the limit, and in case I must solve the last equation, how can I solve it?  Thanks in advance!
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.