# Tag Info

5

$$F_{n+1}=F_{n}+F_{n-1} \Rightarrow \frac{F_{n+1}}{F_n}=1+\frac{F_{n-1}}{F_n}$$ Let $$x_n:= \frac{F_{n+1}}{F_n}$$ Then $$x_n=1+\frac{1}{x_{n-1}}$$ You can now prove that $1 \leq x_n \leq 2$ and by induction that $$x_2 \leq x_4 \leq x_6 \leq .. \leq x_{2n} \leq x_{2n-1} \leq x_{2n-3} \leq .. \leq x_1$$ From here you get that $x_{2n-1}$ and $x_{2n}$ ...

4

Let $R_n=\frac{F_{n+1}}{F_n}$. Since: $$F_n^2-F_{n-1}F_{n+1} = (-1)^n\tag{1}$$ (it is easy to prove by induction) we have that: $$\left|R_{n+1}-R_n\right|=\frac{1}{F_n F_{n+1}}, \tag{2}$$ hence $\{R_n\}_{n\in\mathbb{N}}$ is a Cauchy sequence and $R_n$ converges to some $L>1$ that satisfies $L=1+\frac{1}{L}$, since $R_n>1$ and ...

3

Define $A(x)=\sum_{n\ge 0} a_n x^n$ as the generating function for $(a_n)$. Then, the given recurrence gives you, $$A(x)-(a_1x+a_0)=x(A(x)-a_0)+2x^2A(x)\\\implies A(x)=\frac{-3x}{2x^2+x-1}=-\frac{3x}{(2x-1)(x+1)}=\left(\sum_{n\ge 0}(2x)^n-(-1)^nx^n\right)\\\implies a_n=(2^{n}+(-1)^{n+1}),\ n\ge 1$$

3

Would you be ok with an answer like, i(n) = (2*n+2)*Psi(n+1)+(2*gamma-4)*n+2*gamma; where Psi is the digamma function as defined here? I got that by issuing following in Maple 2015, rsolve( {(it(n+1) - n/(n+1)*it(n) - n + n^2/(n+1) - n/(n+1))*(n+1)/2 = it(n), it(0)=0}, it(n) );

2

Find $\alpha,\beta$ such that $F_0=\alpha+\beta$ and $F_1=\alpha\phi+\beta(1-\phi)$ and show by induction that $$F_n=\alpha\phi^n+\beta(1-\phi)^n.$$ Then compute the desired limit (and use that $|1-\phi|<1$).

2

For (b), the first recurrence implies $$a_{n+2} = 3a_{n+1} + b_{n+1}.$$ Now substitute $a_{n+1} = 3a_n + b_n$ and $b_{n+1} = 5a_n - b_n$ into the right hand side of this equation and regroup to get $$a_{n+2} = 3(3a_n + b_n) + 5a_n - b_n = 14a_n + 2b_n = 8a_n + 2(3a_n + b_n) = 8a_n + 2a_{n+1}.$$

2

This is a really interesting problem! I have only looked at it for a few minutes, so I haven't proved anything, but looking at your triangle array I spotted something interesting: there is a very striking resemblance to Pascal's Triangle!!! If you go by columns, starting with the second column: $1, 2, 3, 4, 5, \cdots$ are all divisible by $1$ and are equal ...

1

Hint: The first term describes the situation in which box $n$ is empty. Term $i$ in the second sum (that should really go from $1$ to $N$) describes the situation in which box $n$ has $i$ balls.

1

Using the $W$ Lambert function (the inverse of $x\mapsto xe^x$) we get: $$a_n=n^n$$ $$\ln a_n=n\ln n=\ln n e^{\ln n}$$ $$W(\ln a_n)=\ln n$$ Therefore $$a_{n+1}=[e^{W(\ln a_n)}+1]^{e^{W(\ln a_n)}+1}$$ On the other hand, from your equation $$(n+1)^{n+1}=n^n\,b^{f(b,n)}$$ we get ...

1

Consider the problem but with the condition that the first and last number is always a $2$. This corresponds to the original problem by chopping of the $2$s: $$2\underbrace{010120}_{\text{original}}2$$ Let $a_n$ be the number of such sequences of length $n$. Then $a_3 = 2$ is the first nonzero entry. Now disregarding that the second last number may not be ...

1

A continuous version of these equations would be $$\frac{d}{dt}\ln a(t)=\ln b(t)\\ \frac{d}{dt}b(t)=a(t)\\ \ddot{b}=\dot{b}\ln b\\ \dot{b}=b\ln b-b+C\\ t=\int\frac{db}{b\ln b-b+C}$$ Since $t\to\infty$ when $b\to1$, we need $C=1$. The leading terms near $b=1$ are $$t=\int \frac2{(b-1)^2}+\frac2{3(b-1)}-\frac19+\frac{7(b-1)}{135}+...db$$ so to leading order ...

1

If you are familiar with homogenic recursions: You can write the characteristic equation: $x^{n+2} = x^{n+1}+2x^n \rightarrow x^2=x+2 \rightarrow x^2-x-2=0$. Your solutions are: $x_1=2, x_2=-1$. Therefore, your recursion have the followning form. $x_n = \alpha * 2^n + \beta *(-1)^n$ Now, to make it correct, we need to get $\alpha$ and $\beta$ with your ...

1

First Step Write the equation in standard form. $$a_{n+2} - a_{n+1} -2a_n = 0$$ Now write the characteristic equation. $$r^2 - r - 2 = 0$$ $\iff r_1 = 2$ & $r_2 = -1$ Thus we know the general form of the homogeneous equation is: $$a_n = c_1{2}^n + c_2({-1})^n$$ Second Step Use the initial conditions. $$a_0 = 0$$ $$a_1 = 3$$ We get: ...

1

It seems the solver is not able to handle the summation statement. Trying now to avoid the summation: \begin{align} i(n) &= n - 1 + \frac{2}{n} \sum_{k=0}^{n-1}i(k) \\ i(n+1) &= n + \frac{2}{n+1} \sum_{k=0}^ni(k) \\ &= \frac{2}{n+1}i(n) + n + \frac{2}{n+1} \sum_{k=0}^{n-1} i(k) \\ &= \frac{2}{n+1}i(n) + n + \frac{n}{n+1} \left( i(n) - n + 1 ...

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