Question:  Show that if $ $ $ k |n \implies F_k |F_n$  $ $  Comments:  I ran into this problem the other day and have not figured it out yet. I think it has something to do with the order of the period of the Fibonacci sequence modulo $m$, for some $ m \in \Bbb{N}$
123714
Question: Let $m,n\in\mathbb{N}$, prove that if $n|m$, $F_n|F_m$.  I've tried to use induction, but I don't really know where to start since there's $2$ numbers:  $n$ and $m\ \dots$ I did induction before with just $1$ number, like proving $1+2+\cdots+n=\frac{n(n+1)}{2}$, that's only dealing with one number $n$, but with this I have no clue how to do it.  Can anyone give me a little hint on how to start, I don't want you to do whole problem for me but can I have some hint so I know where to start?  Thanks for the help!
103599
can anyone help me with this, I got stuck on it.  let $a_n$ and $b_n$ (for $n$ a non-negative integer) be two sequences of real numbers such that $\lim_{n\to \infty}a_n=a$ and $\lim_{n\to \infty}b_n=b$. prove that $\lim_{n\to \infty} \frac{a_0b_n+a_1b_{n-1}+....+a_nb_0}{n}=ab$.
4017
Show that there is no function $f: \mathbb{N} \to \mathbb{N}$ such that $$f(f(n))=n+1987, \ \forall n \in \mathbb{N}$$  This is problem 4 from IMO 1987 - see, for example, .
7520
Let $1\leq p &lt; \infty$. Suppose that    $\{f_k\} \subset L^p$ (the domain here does not necessarily have to be finite),  $f_k \to f$ almost everywhere, and  $\|f_k\|_{L^p} \to \|f\|_{L^p}$.    Why is it the case that $$\|f_k - f\|_{L^p} \to 0?$$   A statement in the other direction (i.e. $\|f_k - f\|_{L^p} \to 0 \Rightarrow \|f_k\|_{L^p} \to \|f\|_{L^p}$ ) follows pretty easily and is the one that I've seen most of the time. I'm not how to show the result above though.
71750
Note that a product   may be parenthesized in two different ways:   and . Similarly, there are several different ways to parenthesize . Two such ways are  and . Let  be the number of different ways to parenthesize the product . Show that if , then      for all integers .  Keep in mind that the sequence  is the same as the sequence of Catalan numbers.   I am entirely lost. Please can someone offer some hint or direction of where to begin? I humbly thank you so very much.
99768
I would like some help with this exercise.  Suppose that $f_1,\ f_2 \in V^*$ and that $\text{Ker} f_1 = \text{Ker} f_2$. Show that $f_1 = k f_2$ for some scalar $k$.  I expect your suggestions.  Thanks!
53269
Let $(x_n)$ be a sequence of real numbers.  Prove that if there exists $x$ such that every subsequence $(x_{n_k})$ of $(x_n)$ has a convergent (sub-)subsequence $(x_{n_{k_l}})$ to $x$, then the original sequence $(x_n)$ itself converges to $x$ .              Thanks for any help.
42409
Came across the following exercise in Bartle's Elements of Real Analysis. This is the solution I came up with. Would be grateful if someone could verify it for me and maybe suggest better/alternate solutions.  I also looked up these related questions - , ,  - but was not happy with proofs given there. I seem to need some help understanding these. Any such help is appreciated.      Show that every sequence in $\Bbb R$ either has a monotone increasing sub-sequence or a monotone decreasing sub-sequence.   Let $(x_n)$ be a sequence in $\Bbb R$. Suppose $(x_n)$ is not bounded. Without loss of generality we may assume that $(x_n)$ is not bounded above. Therefore given any real number there is a member of the sequence which is greater. Let $x_{n_1}$ be any member of the sequence.   There is $x_{n_2} \gt \sup\{x_1, x_2, ..., x_{n_1} \}$. For $i \gt 1$ let $x_{n_i} = \{x_1, x_2, ..., x_{n_{i - 1}}\}$ then $(x_{n_k})$ forms a monotone subsequence of $(x_n)$.   Now suppose instead that $(x_n)$ is bounded. By the Bolzano-Weierstrass Theorem there is a subsequence $(y_n)$ of $(x_n)$ which converges to a limit $y$. Without loss of generality there are infinitely many distinct values in $(y_n)$ that are unequal to $y$. Let $y_{k1}$ be the first such element. Let $y_{k2}$ be any element in $\{ y' \in (y_n) \ \ | \ \ |y' - y | \lt |y - y_{k1}|  \}$.   For $i \gt 1$ let $y_{ki} \in \{ y' \in (y_n)  \ \ | \ \ |y' - y | \lt |y - y_{k \ i - 1}|  \}$. Such $y_{ki}$ exists for every $i \in  \Bbb N$ since $ \lim (y_n) = y $. Now let $(y_{kn})$ be the sub-sequence of $(y_n)$ thus formed. At least one of the two following sets must contain infinitely many elements.    $\{ y \in (y_{kn}) \ \ | \ \ y \gt x\}$ $\{ y \in (y_{kn}) \ \ | \ \ y \lt x\}$   The one which does forms a monotone subsequence.
55194
Suppose that when $n=k$  $(k\geq4)$, we have that $k!&gt;2^k$.  Now, we have to prove that $(k+1)!\geq2^{k+1}$ when $n=(k+1)$ $(k\geq4)$.  $$(k+1)! = (k+1)k! &gt; (k+1)2^k  \text{ (since }k!&gt;2^k)$$  That implies $(k+1)!&gt;2^k2$, since $(k+1)&gt;2$ because of $k$ is greater than or equal to 4.  Therefore $(k+1)!&gt;2^{k+1}$  Finally  we may conclude that $n!&gt;2^n$ for all integers $n\geq4$    Solved, thank you!
39191
Given only $F_n$, that is the $n$th term of the Fibonacci sequence, how can you tell if $n \equiv 1 \mod 2$ or $n \equiv 0 \mod 2$?  I know you can use the Pisano period, however if  $n \equiv 1$ or $n \equiv 2$ $\mod \pi(k)$, it can never be found, where $k$ is in $\pi(k)$ (The Pisano period).  Also there is the fact that if $\sqrt{5F_n^2+ 4}$ is an integer then $n  \equiv 0 \mod 2$, but is there a faster way?  Lastly, because $F_1 = F_2 = 1$, that would have to be an exception for whatever rule/formula that would apply.
91259
Let $q$ be a prime and $n\ge 2$ an integer. Moreover, define $f_q(n)$ as the number of $q$-th residues modulo $n$. Is it true that if $K$ is a positive constant then there exist infinitely many $n$ such that $$ f_q(n)\le Kn? $$    I would like to add that the result holds if $q=2$, indeed it is known that $$ \limsup_{n \to \infty}\frac{f_q(n)}{n}=\frac{1}{2} \,\,\text{ and }\,\,\liminf_{n \to \infty}\frac{f_q(n)}{n}=0. $$
81469
Why is it that in $\mathbb F_q\setminus\{0\}$, when $q\neq 2^k$, there are exactly as many squares as non-squares?  My combinatorics textbook states this without proof :/
52311
Prove that any sequence of real numbers satisfying $a_{n+1}=|a_n|-a_{n-1}$ is periodic.   Although it looks simple, I can't prove this statement...  I tried rewriting the first few terms of the sequence, but nothing interesting showed up...  I'd be mostly interested in hints for this one.
39515
Let $f_n$ be the Fibonacci numbers, defined by $f_0 = 0$ and $f_1 = 1$ and $f_n = f_{n-1} + f_{n-2}$ for all $n \ge 2$.  Show that $f_0 - f_1 + f_2 - \cdots - f_{2n-1} + f_{2n} = f_{2n-1} - 1$ when $n$ is a positive integer.  Just some homework help. Need to prove. Thank you in advance.
103459
Suppose $f:\mathbb{R}\to\mathbb{R}$ has period 1, and for some $q\in(0,1)$:  $$|f(x)-f(y)|\leq q|x-y|\quad \forall x,y$$  Let $g(x)=x+f(x)$, for any $a\in\mathbb{R}$,define the following sequence:  $$a_1=a,\quad a_2=g(a_1),\quad a_3=g(a_2),\quad \dots,\quad a_{n+1}=g(a_n)$$  Show that  $$\lim_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{n^2}$$  exists and does not depend on the choice of $a$.  I don't even know what is the possible approach, any hints? Thanks!
13060
Let $(x_n)_{n\in \mathbb{N}}$ be a sequence of real numbers. A well known result is:     The sequence $(x_n)_{n\in \mathbb{N}}$ converges to $a\in \mathbb{R}$ if and only if every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ converges to $a\in \mathbb{R}$.   Now, I'm working on an exercise which essentialy asks to prove the following:     The sequence $(x_n)_{n\in \mathbb{N}}$ converges to $a\in \mathbb{R}$ if and only if for every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ there is a (sub)subsequence $(x_{n_{k_j}})_{j\in \mathbb{N}}$ which converges to $a$.   Now, one direction is trivial. If in fact $(x_n)_{n\in \mathbb{N}}$ converges to $a$, then by the result I've mentioned, every subsequence of $(x_{n})_{n\in \mathbb{N}}$ converges to $a$. Let then $(x_{n_k})_{k\in \mathbb{N}}$ be an arbitrary subsequence. In that case, we know that it is a subsequence of itself, and hence we've found a (sub)subsequence which converges to $a$.  The other way around seems quite complicated for me. The reason is as follows: I must assume that every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ has a (sub)subsequence which converges to $a$ and the proof that the original one converges to $a$.  My problem is exactly that "a". If it had all (sub)subsequences converging to $a$, then by the result i've mentionted, the subsequence $(x_{n_k})_{k\in \mathbb{N}}$ would converge to $a$. In that case, we would have every subsequence converging to $a$ and the result would follow.  My idea to prove the claim was then:  First remember that a subsequence is the restriction of the original one $x: \mathbb{N}\to \mathbb{R}$ to an infinite subset $\mathbb{N}'\subset \mathbb{N}$. In that case a (sub)subsequence is the restriction of $\tilde{x} = x|\mathbb{N}'$ to an infinite subset $\mathbb{N}''\subset \mathbb{N}$. But since $\tilde{x}$ is the restriction of $x$ to $\mathbb{N}'$ and since $\mathbb{N}''\subset \mathbb{N}'$, the restriction of $\tilde{x}$ to $\mathbb{N}''$ is itself the restriction of $x$ to $\mathbb{N}''$.  In other words, every (sub)subsequence can be viewed as a subsequence of the original one. Considering this, by hypothesis every subsequence has a (sub)subsequence which converges to $a$. But by this argument, this means that we can replace every subsequence with a convergent one. And then we can use the result I stated.  Is my idea correct? Is this really the way to prove the claim? My only doubt is that I need every subsequence converging to $a$ to use the result. In this case I can't see how I can argue that this implies every subsequence converges to $a$.  How can I finish this proof?
26109
Can any one give a generalization of the following properties in a single proof? I have checked the results, which I have given below by trial and error method. I am looking for a general proof, which will cover the all my results below:   Every third Fibonacci number is even. 3 divides every 4th Fibonacci number. 5 divides every 5th Fibonacci number. 4 divides every 6th Fibonacci number. 13 divides every 7th Fibonacci number. 7 divides every 8th Fibonacci number. 17 divides every 9th Fibonacci number. 11 divides every 10th Fibonacci number. 6, 9, 12 and 16 divides every 12th Fibonacci number. 29 divides every 14th Fibonacci number. 10 and 61 divides every 15th Fibonacci number. 15 divides every 20th Fibonacci number.