Let $(u_n)$ be a real sequence defined $u_0&gt;0$ and $u_{n+1} =\dfrac{1}{2} \arctan(u_n)$. We define $v_n = 2^n u_n$ and write $l = \lim v_n$.  Find an equivalent of $u_n - \dfrac{l}{2^n}$ as $n\to \infty$.  What I did:  I've shown that $(u_n)$ converges to 0, $(v_n)$ is convergent. So we get $u_n \sim \dfrac{l}{2^n}$. I don't know how to continue.
120741
(I'm aware of  but it has no answers…)  Let be $u_0 \in \mathbb{R}$ and the sequence $(u_n)_n$ defined by: $u_{n + 1} = \frac12 \arctan(u_n)$.  I define also: $v_n = 2^n u_n$, so I can show that: $\lim (u_n)_n = 0$ (by studying $x \mapsto \frac12 \arctan(x)$), thus, I can show that $(v_n)_n$ is monotone and converges because it is bound.  Now, I conclude: $u_n \sim \dfrac{l}{2^n}$, I'd like to determine $l$ more precisely.  Here is what I tried, I suspect $l$ to be something like $f(\pi)$ for some $f$ :   push the asymptotic expansion of $\arctan$ to the 2nd order and reinject it ; use $\arctan(u_n) + \arctan(1/u_n) = \dfrac{\pi}{2}$ ; use series techniques to look for $\sum v_{n + 1} - v_n$, maybe conclude using Cesaro summation
42955
Given a sequence $a_{n}$, if I know that the sequence of even terms converges to the same limit as the subsequence of odd terms:    $$\lim_{n\rightarrow\infty} a_{2n}=\lim_{n\to\infty} a_{2n-1}=L$$  Is this sufficient to prove that the $\lim_{n\to\infty}a_{n}=L$?    If so, how can I make this more rigorous?  Is there a theorem I can state that covers this case?
1309
Show that $U_n:=\left(1+\dfrac{1}{n}\right)^n$, $n\in\Bbb N$, defines a monotonically increasing sequence.   I must show that $U_{n+1}-U_n\geq0$, i.e. $$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\geq0.$$  I am trying to go ahead of this step.
52746
Does the condition $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ imply that $a_n$ convergent?  When I say convergent I mean to $L\in \mathbb{R}$ or to $\infty$ or to $-\infty$  I have a gut feeling this statement is false, just can't find a series which contradicts it.  Thanks a lot
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?
84905
I need some verification for my proof in part a) and help to get me started on part b)  a) Prove that the sequence $a_n = (2n+1)/(3n+5)$ converges to $2/3$ directly from the definition of convergence of a sequence.  Solution: Some rough work first: Given $\epsilon&gt;0$, we want $|a_n-(2/3)|&lt; \epsilon$. Need $|(2n+1)/(3n+5) - (2/3)|&lt; \epsilon$ which after simplifying gives $7/(9n+15)&lt; \epsilon$. Solving for $n$ gives $n&gt; (7-15\epsilon)/9\epsilon$.  Then we write our formal proof: Given $\epsilon&gt;0$, set $N= (7-15\epsilon)/9\epsilon$. Let $n&gt;N$. Then $|a_n-(2/3)| = 7/(9n+15) &lt; 7/(9((7-15\epsilon)/9\epsilon)+15) =\epsilon$.  Thus by definition the sequence converges to 2/3.  b) Suppose that the sequence $\{a_n\}$ is not bounded above (that is, for any real number $x$, there is some $n$ so that $a_n&gt;x$). Prove that $\{a_n\}$ does not converge to $42$.  To solve this I am assuming that we have to suppose that $\{a_n\}$ does converge to $42$, and then we will get a contradiction, that is the sequence will be bounded above. So can anyone help me get started on this?
15022
Here's the original question:     Let $(a_n)$ be bounded. Assume that $a_{n+1} \ge a_{n} - 2^{-n}$. Show   that $(a_n)$ is convergent.   Okay, I know that if I can show that if the sequence is monotone, I can conclude that it is convergent. But I am not sure how to show that it is monotone.  I know that  $$a_n \le a_{n+1} + \frac{1}{2^n} &lt; a_{n+1} + \frac{1}{n}$$  It looks to me as if it is monotonically increasing but I'm quite not sure how to prove my claim. Any hints would be appreciated.
16916
Let $\{a_n\}$ be a sequence of real numbers such that $|a_{n+1} - a_n| \le \dfrac{1}{2^n}$.   I would like to show that this sequence is Cauchy.    Letting $\epsilon &gt; 0$, I said choose $N$ such that $1/2^N \le \epsilon$. However, I'm not sure if this is the right $N$, since $m, n &gt; N$ doesn't seem to necessarily imply $|a_n - a_m| \le \epsilon$.  Can someone show me the light?
122166
Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of Riemann-integrable functions on $[a, b]$.   Assume that $(f_n)_{n\in\mathbb{N}}$ converges pointwise to $f$ on $[a, b]$ and converges in the mean to $g$ on $[a, b]$, where $f$ and $g$ are both continuous. Show that $f = g$.   ($(f_n)_{n\in\mathbb{N}}$ converges in the mean to g on $[a , b]$ means that $\lim_{n\to\infty} d(f_n, g) = 0$, where $d$ is the $L_2$ metric.)  I am studying real analysis now and want to prove this statement. But I don't know how to use condition "$(f_n)_{n\in\mathbb{N}}$ converges in the mean to g on $[a , b]$." How can I prove this?
53291
Problem  The sequence $(a_n)_{n=1}^\infty$ is given by recurrence relation:   $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$.   Evaluate the limit $\lim_{n\to\infty} a_n$.  Solution   Show that the sequence $(a_n)_{n=1}^\infty$ is monotonic. The statement $$V(n): a_n &lt; a_{n+1}$$ holds for $n = 1$, that is $\sqrt2 &lt; \sqrt{2+\sqrt2}$. Let us assume the statement holds for $n$ and show that $V(n) \implies V(n+1)$. We have that $$a_n &lt; a_{n+1}.$$ Adding 2 to both sides and taking square roots, we have that $$\sqrt{2+a_n} &lt; \sqrt{2+a_{n+1}},$$ that is $a_{n+1} &lt; a_{n+2}$ by definition. Find bounds for $a_n$. The statement $$W(n): 0 &lt; a_n &lt; 2$$ holds for $n=1$, that is $0 &lt; \sqrt2 &lt; 2$. Let us assume the statement holds for $n$ and show that $W(n) \implies W(n+1)$. We have that $$0 &lt; a_n &lt; 2.$$ Adding two and taking square roots, we have that $$0 &lt; \sqrt2 &lt; \sqrt{2+a_n} &lt; \sqrt4 = 2.$$ The limit $\lim_{n\to\infty} a_n$ exists, because $(a_n)_{n=1}^\infty$ is a bounded monotonic sequence. Let $A = \lim_{n\to\infty} a_n$. Therefore the limit $\lim_{n \to\infty} a_{n+1}$ exists as well and $\lim_{n \to\infty} a_{n+1} = A$. (For $(n_k)_{k=1}^\infty = (2,3,4, \dots)$, we have that $(a_{n_k})_{k=1}^\infty$ is a subsequence of $(a_n)_{n=1}^\infty$, from which the statement follows.) We have that $a_{n+1} = f(a_n)$. That means that $A = \lim_{n\to\infty} a_n = \lim_{n \to\infty} {f(a_n)} = f(\lim_{n \to\infty} a_n) = f(A) = \sqrt{2 + A}$. Solving the equation $A = \sqrt{2 + A}$, we get $A = -1 \lor A = 2$. Putting it all together, we get that $A = 2$, because the terms of the sequence are increasing and $a_1 &gt; 0$.   Is my solution correct?
79021
I am looking for an example of a sequence which shows that, $a_{n+1} - a_n \to 0$ as $n \to \infty$ does not imply that sequence $a_n$ converges. I have a feeling that the sequence should be an oscillatory one but I am unable to think of an example.
98908
Let $(f_n)$- real sequence such that $$ \lim_{n\rightarrow \infty} \frac{f_{n+1}f_n-f_{n-1}f_{n+2}}{f_{n+1}^2-f_nf_{n+2}}=a+b, $$ and $$ \lim_{n\rightarrow \infty} \frac{f_{n}^2-f_{n-1}f_{n+1}}{f_{n+1}^2-f_nf_{n+2}}=ab   \quad    (|a|&lt;|b|). $$  Prove that:$$\lim_{n\rightarrow \infty} \frac{f_n}{f_{n+1}}=a $$  I think we must prove $\displaystyle\lim_{n\rightarrow \infty} \frac{f_n}{f_{n+1}} $ exists,and we prove this limit is $a$,But I can't prove this limit exists.  My idea: since $$\lim_{n\to\infty}\dfrac{\dfrac{f_{n}}{f_{n+1}}-\dfrac{f_{n-1}}{f_{n}}\dfrac{f_{n}}{f_{n+1}}\dfrac{f_{n+2}}{f_{n+1}}}{1-\dfrac{f_{n}}{f_{n+1}}\dfrac{f_{n+2}}{f_{n+1}}}=a+b$$ and  $$\lim_{n\to\infty}\dfrac{\left(\dfrac{f_{n}}{f_{n+1}}\right)^2-\dfrac{f_{n-1}}{f_{n}}\dfrac{f_{n}}{f_{n+1}}}{1-\dfrac{f_{n}}{f_{n+1}}\dfrac{f_{n+2}}{f_{n+1}}}=ab$$ But I felt this deal is not useful,  Other idea: I want to take the  Fibonacci sequence  to solve this problem, But I can't,   Thank you
83427
I have been trying to solve this problem for awhile, and I can't seem to solve it.Below is the question. I suspect that the series does converge, but I am unable to prove this. Thanks for any help in advance.  Let $(a_n)_{n=1}^\infty$ be a convergent sequence of real numbers. Can it be concluded that $\sum\limits_{n=1}^{\infty}|a_{n+1} - a_{n}|$ converges? Give a proof or a counterexample.
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!
115748
The sequence is defined: $$u_1=1, u_{n+1}=1+\dfrac n {u_n}$$  The question asks to find an asymptotic development of 2 terms for $n\to+\infty$. I have got $u_n\sim_\infty\sqrt{n} $, but how to derive the 2nd term?
44461
I would appreciate your help!  How can we show that if a sequence of positive real numbers $a_n$ converges to $a\in\mathbb{R}$ with $a&gt;0$, then $(a_0 a_1\cdots a_n)^{\frac{1}{n}}$ converges to $a$.
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.