We define the $n^{th}$ Fibonacci number as $a_1 = a_2 = 1$ and $a_n = a_{n-1} + a_{n-2}$ for $n \geq 3$. Consider  $$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n}. $$  I wrote a script and found that this limit converges to the golden ratio $\phi \approx 1.61803$. However, I'm having trouble giving a rigorous proof for this. Here's what I have so far:  $$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{a_n + a_{n-1}}{a_n} = \lim_{n \to \infty} \left( 1 + \frac{a_{n-1}}{a_n} \right). $$  I'm unsure of how to proceed after this. I know that expressing the limit as a certain continued fraction would imply that it converges to $\phi$ but I'm unsure of how to get there. Any ideas?
31125
How would one prove that $$\lim_{n\rightarrow \infty} \frac{F_{n+1}}{F_n}=\frac{\sqrt{5}+1}{2}=\varphi$$  where $F_n$ is the nth Fibonacci number and $\varphi$ is the Golden Ratio?
66113
Let $\ a_n=\sqrt{n+1}-\sqrt{n}$. I have to show that $\lim_{n\to \infty}a_{n}=0$.     How should I start? Do I have to use any theorem?
27798
Question: Let $F_n$ the sequence of Fibonacci numbers, given by $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$. Show for $n, m \in \mathbb{N}$: $$F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$$   My (very limited) attempt so far: after creating a small list of the values $F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21, F_9=34, F_{10}=55$ i can see that yes it does seem to work for instance $F_{6+3}=F_5 F_3 +F_6 F_4 = 10 +24 = 34 = F_9$. However, I really don't know where to begin as showing that this must hold in general terms. Should I be looking to use limits? Or perhaps induction? What is the best way to solve this?
3243
For any $a &gt; 0$, I have to show the sequence $x_{n+1}$ $=$  $ \frac 12$($x_n+ $ $ \frac {a} {x_n}$)  converges to the square root of $a$ for any $x_1&gt;0$.  If I assume the limit exists ( denoted by $x$) then,  $x$ $=$  $ \frac 12$($x+ $ $ \frac {a} {x}$) can be solved to $x^2 = a$  How could I show that it does exist?
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 $\sqrt{3}$?
108876
Let a sequence be $a_1=1;a_{n+1}=(n+1)(1+a_n)$ If $P_n=\prod_1^n(1+a_i^{-1})$ then $$\lim_{n\to\infty}P_n $$is ?    I did: $$P_n=\prod_1^n\frac{(1+a_i)}{a_i} =\frac{a_{n+1}}{a_1}\prod1^{n-1}\frac{(1+a_i)}{a_{i+1}} =\frac{a_{n+1}}{a_1}\prod_1^{n-1}\frac1{i+1}=\frac{a_{n+1}}{n!}$$ Now how do I find the limit?
3339
If $F_n$ is the $n$-th fibonacci number, then prove that, $$F_n={n-1 \choose 0 }+{n-2 \choose 1 }+{n-3 \choose 2 }+\ldots$$  I tried the idea of using Pascal's triangle, but it seems to need some adjustment, so I am stuck, please help.    Any combinatorial argument would be of greater help than rigorous proof.  Thank you.
68075
Problem: Find with proof limit of the sequence $a_n=1+\frac{1}{a_{n-1}}$ with  $a_1=1$  or show that the limit does not exist.  My attempt:  I have failed to determine the existence. However if the limit exists then it is easy to find it.  The sequence is not  monotonic and I have failed to find any monotonic subsequence subsequence. The sequence is clearly bounded below by $1$. I have observed that the sequence is a  continued fraction so it alternatively increases and decreases.   So, please help me.
23624
Let $ a_{n+1} = \frac{1}{3}(2a_n + \frac{5}{a_n^{2}}), a_1 &gt; 0 $. Show the sequence converge and find it's limit.    I did the following:  We see that $a_n$ is positive for every n obviously (considering $a_1 &gt; 0$). Then:  $$ a_{n+1} - a_n = \frac{2a_n^{3} + 5}{3a_n^{2}} - a_n = \frac{5 - a_n^{3}}{3a_n^{2}}$$  We see that for each $a_n &gt; ^3\sqrt{5}$ the sequence is decreasing. That means the proof collapses to prove  $a_n &gt; ^3\sqrt{5}$ in order for $a_n$ to be decreasing. Suppose $ a_n &gt; ^3\sqrt{5}$ then obviously $a_{n+1} = \frac{2a_n^{3} + 5}{3a_n^{2}} &gt; \frac{2(\sqrt[3]{5})^{3} + 5}{3(\sqrt[3]{5})^{2}} = \sqrt[3]{5}$ so we deduced $a_{n+1} &gt; ^\sqrt[3]{5} $.  (It's important to note $a_1$ might be less then $\sqrt[3]{5}$, hence the sequence will decrease from the second element.) So we observed $a_n$ is always positive and decreasig hence it's bounded below so it's bounded. As for it's limit:   $$\begin{align} L = \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \frac{1}{3}(2a_n + \frac{5}{a_n^{2}}) = \frac{1}{3}(2L + \frac{5}{L^{2}}) =  \sqrt[3]{5}\end{align}$$  I'm not sure as of the correctness of the solution. I'm aware that it's not a contradiction because it only implies it's in the epsilon-environment of $\sqrt[3]{5}$, or rather in this case $(\sqrt[3]{5}, \sqrt[3]{5}+\epsilon)$.  Is it correct?  EDIT: If anyone encounter a question in the likes, to sum it up you obviously should show the sequence is montonic and bounded. In this case for example we can show it using AM-GM inequality to see the sequence is bounded by $\sqrt[3]{5}$ and then in the difference of two consequective elements we see the difference is negative due to every element being greater then the third root of 5. From here we know the sequence is bounded below and decreasing hence it's bounded. If we plug in the limit of the sequence to itself we will see it's the third root of five. And that's it.
81098
The sequence $a_n=(1/2)^{(1/3)^{...^{(1/n)}}}$ doesn't converge, but instead has two limits, for $a_{2n}$ and one for $a_{2n+1}$ (calculated by computer - they fluctuate by about 0.3 at around 0.67). Why is this?
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 "."
108257
Find the limit of the sequence $a_n=\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}.$  The limit is supposed to be $e^{-1}$ and there are a couple of posts in MSE proving it. But here is a proof I encountered showing the limit is $1.$ Please tell me what's going wrong here.      Proof:  We can write $a_n=\sqrt[n]{n!}(b_n-1),$ where $b_n=\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}.$ Hence $$a_n=n\cdot\frac{\sqrt[n]{n!}}{n}\cdot\frac{b_n-1}{\ln{b_n}}=\frac{\sqrt[n]{n!}}{n}\cdot\frac{b_n-1}{\ln{b_n}}\cdot\ln{b_n^n.}$$   But $\lim \limits_{n\to \infty}\frac{\sqrt[n]{n!}}{n}=e^{-1},$ so $b_n\to1$ as $n\to \infty.$ On the other hand, $$\lim \limits_{n\to \infty}b_n^n=\lim \limits_{n\to \infty}\frac{(n+1)!}{n!}.\frac{1}{\sqrt[n+1]{(n+1)!}}=e.$$   Also $$\lim \limits_{n\to \infty}\frac{b_n-1}{\ln{b_n}}=\lim \limits_{n\to \infty}\frac{1}{\ln{[1+(b_n-1)]^{1/(b_n-1)}}}=\frac{1}{\ln e}=1$$         And the author goes on to say that the limit is $e^{-1}.$ But doesn't it say that the limit is $1?$ I must have overlooked something so obvious.
48910
So I'm being asked to find a formula for the nth fibonacci number.    I know the answer is $$x_{n}=\frac{(1+5^{1/2})^{n} -(1-5^{1/2})^n}{\sqrt{5}2^n}$$  However I don't really know how to get there.  It seems so complicated and I need some help.
86577
How is a continuous Fibonacci number $F_n$ defined? (like the Gamma function for factorials) Tried with Binet's formula (Wiki) , but to no luck. F[n_] = N[GoldenRatio^n - 1/GoldenRatio^n]/Sqrt[5] Plot[F[n], {n, 3., 12.}, GridLines -&gt; Automatic] FibonRatio[n_] = N[Fibonacci[n + 1]/Fibonacci[n]]; Plot[{FibonRatio[n], N[GoldenRatio], 1.7}, {n, 3., 12.},   GridLines -&gt; Automatic]  Also using this continuous function definition how is it proved that $$\lim _{n\rightarrow \infty}\dfrac{F_{n+1}}{F_n} = \phi$$ ( where $\phi$ is the GoldenRatio) ? Appreciate your comments.
106593
Please help with my homework question:  Define $\{a_n\}_{n=1}^{\infty}$, $\{b_n\}_{n=1}^{\infty}$ as:  $b_{n+1}=\frac{a_n+b_n}{2}, a_{n+1}=\frac{a_n^2+b_n^2}{a_n+b_n}, 0&lt;b_1&lt;a_1$  Prove the limits converge and that $\lim \limits_{n \to \infty}a_n = \lim \limits_{n \to \infty}b_n $.  I have tried to show that they are monotone and bounded, tried looking for a property of arithmetic mean (related to $b_n$) but I'm completely stuck as none of these yielded anything useful.  Any help would be appreciated
28088
I have the following problem.   I have to find the limit of $b_n = \frac{a_n}{n}$, where $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=l$  My approach:  I express $a_n$ in terms of $b_n$, i.e. $a_n=nb_n$ and $a_{n+1}=(n+1)b_n$  We look at the difference: $a_{n+1}-a_n=(n+1)b_{n+1}-nb_n$  Assuming that $b_n$ converges to a real number m, we see that:  $l=(n+1)m-nm$, from where I conclude that $m=l$.  What I'm left with is proving that $b_n$ is convergent which I'm not sure how to do.  Thanks in advance!
59709
How do I prove that the limit of the sequence exists: $$a_n = \frac1n +\frac 1{n+1} + \frac1{n+2} + \dots + \frac1{2n}$$  I can find the limit as n goes to infinity, $\frac1{2n}$ goes to $0$, so the limit is $0$ right? But how do I prove that the limit exists? Do I have to use the $\varepsilon$ $\delta$ definition of a limit?    For part b) I have to show that the limit is less then $1$ but not less then $\frac12$, I thought the limit was $0$?