$a_1= \sqrt3 $ , all $n\geqq1$, $a_{n+1}=\sqrt{3+a_n}$, prove convergence of sequence ${a_n}$   How to prove... I don't know..
3167
(Fitzpatrick Advanced Calculus 2e, Sec. 2.4 #12)  For $c \gt 0$, consider the quadratic equation $x^2 - x - c = 0, x &gt; 0$.  Define the sequence $\{x_n\}$ recursively by fixing $|x_1| \lt c$ and then, if $n$ is an index for which $x_n$ has been defined, defining  $$x_{n+1} = \sqrt{c+x_n}$$  Prove that the sequence $\{x_n\}$ converges monotonically to the solution of the above equation.  Note: The answers below might assume $x_1 \gt 0$, but they still work, as we have $x_3 \gt 0$.    This is being repurposed in an effort to cut down on duplicates, see here: .  and here: .
65053
let sequence $\{a_{n}\}$ such $$a_{1}=1,a_{n+1}=1+\dfrac{n}{a_{n}}$$  show that: $$a_{n}=\sqrt{n}+\dfrac{1}{2}-\dfrac{1}{8\sqrt{n}}+o\left(\dfrac{1}{\sqrt{n}}\right)?$$
91082
Let a > 0 and let $x_1 &gt; 0$. Define $x_{n+1} := \sqrt{a + x_n}$ for n $\in$ N. Show that &lt;$x_n$> converges and find its limit.
112089
Let $(a_n)$ be a sequence such that $(a_n+a_{n+1})$ is convergent. Prove that $(a_n/n)$ converges to $0$.   $(a_n+a_{n+1})$ convergent to $L$ means that for all $\epsilon$, there exists $N$ such that for all $n\geq N$, $|L-a_n-a_{n+1}|&lt;\epsilon$, or in other words $L-\epsilon&lt;a_n+a_{n+1}&lt;L+\epsilon$. How to proceed from here?
23637
I am trying to find out if the following sequence $(a_n)_{n\geq 1}$ will either converge or diverge. $$ a_n = (-1)^n + \frac1n $$  If the limit converges, I need to find its limit.   I tried plugging in various numbers for n besides 1, but I am not entirely sure if it really does converge.
55354
Ch.3 #7. Prove that the convergence of $\sum a_n$ implies the convergence of $$\sum \frac{\sqrt{a_n}}{n},$$ if $a_n \geq 0$.      My attempt. If $\sum a_n$ is convergent, then by the root test, $\lim_{n \rightarrow \infty} sup \sqrt[n]{|a_n|} &lt; 1$. This means $|a_n|^{1/n} &lt; 1$. Because $a_n \geq 0$, this means $a_n ^{1/n} &lt; 1$. Now apply the root test to $\sum \frac{\sqrt{a_n}}{n}$ by considering $$\lim_{n \to \infty} sup \sqrt[n]{|\frac{\sqrt{a_n}}{n}|}.$$ $$\sqrt[n]{|\frac{\sqrt{a_n}}{n}|} = \frac{\sqrt{{a_n}^{\frac{1}{n}}}}{\sqrt[n]n}.$$ Because  $a_n ^{1/n} &lt; 1$, certainly the numerator on the left-hand side of the equation above is $&lt;1$. And because $n$ tends to $\infty$, and $n &gt; 1 \rightarrow \sqrt[n]n &gt; 1,$ the denominator is $&gt;1$. Therefore the limit is $&lt;1$ and the root test implies that $\sum \frac{\sqrt{a_n}}{n}$ converges.     If there are any errors in my proof, I think they come from 1) my assumption of the converse of the root test theorem and 2) my misunderstanding of $\lim_{n\to\infty} sup$. Thanks in advance for your help.
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.
26008
Suppose $\sum{a_n}$ converges. How do I prove that $\sum{\frac{\sqrt{a_n}}{n}}$ converges.  So I know that just because  $\sum{a_n}$ converges, do not mean I can say anything about the converges of its square root. So I know that I can prove that if  $\sum{\sqrt{a_n}}$ converges then $\sum{\frac{\sqrt{a_n}}{n}}$ converges by Abel's Test, but I do not know where to start for the case where  $\sum{\frac{\sqrt{a_n}}{n}}$ diverges
63598
If $a_n$ is a sequence such that   $$a_1 \leq a_2 \leq a_3 \leq \dotsb$$   and has the property that $a_{n+1}-a_n \to 0$, then can we conclude that $a_n$ is convergent?   I know that without the condition that the sequence is increasing, this is not true, as we could consider the sequence given in  to a similar question that does not require the sequence to be increasing.  $$0, 1, \frac12, 0, \frac13, \frac23, 1, \frac34, \frac12, \frac14, 0, \frac15, \frac25, \frac35, \frac45, 1, \dotsc$$  This oscillates between $0$ and $1$, while the difference of consecutive terms approaches $0$ since the difference is always of the form $\pm\frac1m$ and $m$ increases the further we go in this sequence.    So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge?  Or is this still not sufficient?
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How do I prove that if $|a_{n+1}|/|a_n| \leq |b_{n+1}|/|{b_n}|$ and $\sum |b_n|$ is convergent, then $\sum |a_n|$ is convergent? I know that if I can get $|b_{n+1}/{b_n}| &lt; 1$ then I can use the ratio test to say $\sum a_n$ is convergent, but I don't know if this the proper way to approach the problem.
15680
Does $\sum_{n=1}^{\infty} \sqrt[n]{3} - 1$ converge or diverge?   Hints please  I have that $0 \le \sqrt[n]{3}  - 1 \le 1$  Does this help?
1293
I need help to solve following problem from Rudin's Mathematical analysis book:    Convergence of the series $\sum a_n$  implies the convergence of $\sum \dfrac{\sqrt {a_n}}{n}$, if $a_n&gt;0$  I tried to construct a suitable convergence sequence $b_n$ such that $\sum b_n$ converges and $a_n \leq b_n$ but, I am not able to find such sequence $b_n$ .  Thanks for the help and sugestions.
26570
The last problem I posted had a wrong statement. I recovered the correct one.  Let $(a_n)$ be a sequence of positive real numbers. Prove that $\sum_{n\geq 0}a_n$ converges iff $\displaystyle \sum_{n\geq 0} \frac{a_n}{\sum_{k=0}^n a_k}$ converges  The direct statement is easy to prove using comparison test. I'm (again!) stuck with the converse. I tried summation by part, without success. Note that the convergence of $\displaystyle \sum_{n\geq 0} \frac{a_n}{\sum_{k=0}^n a_k}$ implies $\displaystyle \frac{a_n}{\sum_{k=0}^n a_k} \to 0$ I don't know what to do next ...
61480
We define $a_1=1$ and $a_{n+1}=\dfrac{1}{2}\left(a_n+\dfrac{2}{a_n}\right)$.  I want to prove that $a_n$ converges. But first I'd like to show that $a_n&lt;\sqrt{2}$ for every $n\in\mathbb{N}$.  I try to use induction. For a fixed $n$, I suppose that $a_n&lt;\sqrt{2}$.  Any hint to prove that $a_{n+1}=\dfrac{1}{2}\left(a_n+\dfrac{2}{a_n}\right)&lt;\sqrt{2}$?  Thanks.
120686
Consider the sequence defined by $a_1 = \sqrt{2}$, $a_2 = \sqrt{2 + \sqrt{2}}$, so that in general, $a_n = \sqrt{2 + a_{n - 1}}$ for $n &gt; 1$. I know 2 is an upper bound of this sequence (I proved this by induction). Is there a way to show that this sequence converges to 2? What I think is that the key step is to prove 2 is the least upper bound of this sequence. But how?
15457
Let $a_1=1$ define $a_{n+1}=\frac{1}{2}(a_n+\frac{2}{a_n})$ . Show that the sequence converges to $\sqrt{2}$. The idea is to show the sequence is bounded and monotone . But how should I do it?
30401
I want to prove the following theorem     Let $\{a_{n}\}$ be a convergent sequence. Prove that $\lim\limits_{n\to \infty}a_{n+1}=\lim\limits_{n\to \infty}a_{n}$   Do I need to show that it is monotone increasing and then use the relation  $$0\leq \lim_{n\to \infty}a_{n+1}- \lim_{n\to \infty}a_{n}\leq 0 \;?$$  Please, I need help on this! Various proofs are welcome! Thank you very much for your time!