As $\lim_{m \to \infty}$ , the next number in the sequence would be the sum of all the numbers before it, so the first non-zero numbers of $F$ are $1$ , then $0 + 1 = 1$ , then $0 + 1 + 1 = 2$ , then $0 + 1 + 1 + 2 = 4$ , then $0 + 1 + 1 + 2 + 4 = 8$ , and so on, for the sequence $1, 1, 2, 4, 8, \dots$

After the first $1$ , we observe that the numbers after it are successive powers of $2$ . In other words, if $G_p$ is the sequence of $F_n$ starting at the second $1$ so that $G_0 = 1$ , $G_1 = 2$ , $G_2 = 4$ , and so on, then $G_p = 2^p$ . This observation can be proved inductively: $G_0 = 2^0 = 1$ , and assuming $G_p = 2^p$ , $G_{p + 1} = 1 + {\displaystyle \sum_{k = 0}^{p}2^k} = 1 + \frac{1- 2^{p + 1}}{1 - 2} = 2^{p + 1}$ . Therefore,

$\lim_{m \to \infty}{\phi_m} = \lim_{m \to \infty}{\lim_{n \to \infty}{\frac{F_{n+1}^m}{F_{n}^m}}} = \lim_{p \to \infty}{\frac{G_{p+1}}{G_p}} = \lim_{p \to \infty}{\frac{2^{p+1}}{2^p}} = \boxed{2}$

By the definition of the sequence, we have $F_{n}^m+F_{n+1}^m+F_{n+2}^m+\cdots F_{n+m-1}^m=F_{n+m}^m$ For large enough $n$ , this becomes $1+\phi_m+\phi_m^2+\cdots \phi_m^{m-1}=\phi_m^m \implies \frac{\phi_m^m-1}{\phi_m-1}=\phi_m^m \implies \phi_m^m(\phi_m-2)=-1$ As $m$ goes to infinity, $\phi_m^m$ goes to infinity, so in order for the product to converge, $\phi_m-2$ must tend to 0, therefore the answer is 2.

Assuming $\phi_m$ exists for every $m$ ...

We can divide both sides of $F_n^m = \sum_{i=1}^mF_{n-i}^m$ by $F_{n-m}^m$ and take the limit as $n\to\infty$ to find $\phi_m^m = \sum_{i=1}^{m} \phi_m^{m-i} = \frac{\phi_m^m - 1}{\phi_m-1} \quad \implies \quad \phi_m^m(\phi_m - 2) + 1 = 0$ so $\phi_m$ is a root of $x^m(x-2)+1 = 0.$

Since for each $m$ , the sequence $n \mapsto F_n^m$ is increasing, we must have $\phi_m \geq 1.$ Further, if we look at the first equation I gave for $\phi_m,$ we can see that $\phi_m=1 \implies 1 = \phi_m^m = \sum_{i=1}^m \phi_m^{m-i} = m$ so, since we're only interested in $\lim_{m\to\infty} \phi_m$ , we may assume $\phi_m > 1$ . Further, since for $x\geq 2,$ we have $x^m(x-2)+1 \geq 1 > 0,$ we also have $\phi_m < 2$ for all $m.$

So far, we have shown that if $m>1$ , then $\phi_m \in (1,2)$ is a root of $f_m(x) = x^m(x-2)+1.$ To strengthen this, note that $x=1$ is a root of $f_m$ and $f_m'(x) = (m+1)x^m - 2mx^{m-1} = x^{m-1}\Big((m+1)x - 2m\Big)$ has only one root $x = \frac{2m}{m+1} \in (1,2),$ so by Rolle's theorem (or more generally the Mean Value Theorem) we can conclude that $\phi_m$ is the unique root $\in (1,2)$ of $f_m.$

To finish the proof, we note that for any $0 < \varepsilon < 1,$ $\begin{aligned} m > -\log_{2-\varepsilon}(\varepsilon) &\implies f_m(2-\varepsilon) = (2-\varepsilon)^m(-\varepsilon) + 1 < 0 \\ &\implies f_m \text{ has a root } \in (2-\varepsilon, 2) \subset (1,2) & \text{(Intermediate Value theorem)}\\ &\implies 2-\varepsilon < \phi_m < 2 & \text{(Since } \phi_m \text{ is the unique root } \in (1,2)\text{)} \end{aligned}$

Therefore, we have shown by the formal definition of the limit that $\lim_{m\to\infty} \phi_m = \boxed{2}$

I'm going to play a little fast and loose with limits towards infinity here.

Extend this generalized sequence to a very high value of $m$ . The first terms are a bunch of zeroes then the one but then comes $[...0,0,0,1,] 1, 2, 4, 8, 16, 32,...$ . This doubling sequence will occur for $m$ terms. If we make $m \rightarrow \infty$ then the ratio between successive terms is $\boxed{2}$ .

The generating function for the series is $F(z) = z + zF(z)+z^2F(Z)+...+z^mF(z)$ rearranging the terms $\displaystyle\therefore F(z) = \frac{z}{1-\displaystyle\sum_{i=1}^{m}z^i}$

as $\;\;\;m\rightarrow\infty$ , $\;\;\;\;\displaystyle\lim_{m \to \infty}\sum_{i=0}^{m}z^i =\frac{1}{1-z}$ $\therefore F(z)= \frac{z^2-z}{2z-1}$ $\therefore F(z) = \frac{z}{2}-\frac{1}{4} + \frac{1}{4}\cdot\frac{1}{1-2z}$ $\therefore F(z) = \frac{z}{2}-\frac{1}{4} +\frac{1}{4}\cdot \displaystyle\sum_{i=0}^{\infty}(2z)^i$ $\therefore \lim_{n \to \infty} \phi_{m} = 2$

Since the text gives us that $F^m_n$ = $\displaystyle \sum_{i=1}^m$ $F^m_{n-i}$ we can evaluate $F^m_{n+1}$ as follows: $F^m_{n+1}$ = $\displaystyle \sum_{i=1}^m$ $F^m_{n+1-i}$ = $F^m_n$ + $F^m_{n-1}$ + $\ldots$ + $F^m_{n-m+1}$ = $F^m_n$ + $F^m_{n-1}$ + $\ldots$ + $F^m_{n-m+1}$ + $F^m_{n-m}$ - $F^m_{n-m}$ = $2\cdot F^m_n$ - $F^m_{n-m}$ . It's clear that $\frac {F^m_{n+1}}{F^m_n}$ = $\frac {2\cdot F^m_{n} - F^m_{n-m}}{F^m_n}$ = 2 -- $\frac {F^m_{n - m}}{F^m_n}$ . Since Fibonacci - like sequences, obviously, grow monotonically, for n sufficient large (i.e. for n approaching infinity) the fraction becomes 0 and the result we find is $\boxed {2}$

For $F^m$ , each term (except for the first m ones) is equal to the sum of m previous terms. So if we expand two adjacent terms $F^m_k$ and $F^m_{k+1}$ we get:

$F^m_k=\Sigma_{i=k-m}^{k-1} F^m_i$ and

$F^m_{k+1}=\Sigma_{i=k-m+1}^{k} F^m_i$ .

Using the overlap in the ranges: $F^m_{k+1}=\Sigma_{i=k-m}^{k-1} F^m_i-F^m_{k-m}+F^m_k=2F^m_k-F^m_{k-m}$ .

The ratio between adjacent terms then is $\frac{F^m_{k+1}}{F^m_k}=2-\frac{F^m_{k-m}}{F^m_{k}}$ .

We are told that $\phi_m=\lim_{k\to\infty}{\frac{F^m_{k+1}}{F^m_{k}}}$ exists. From this we can infer that $\lim_{k\to\infty}\frac{F^m_{k-m}}{F^m_{k}}=\phi_m^{-m}$ . And since for m>1, $\phi_m$ must be larger than 1(*), we know $\lim_{m\to\infty}\phi_m^{-m}=0$ , and $\lim_{m\to\infty}\phi_m=\lim_{m\to\infty}\lim_{k\to\infty}\frac{F^m_{k+1}}{F^m_{k}}=2-0$ .

(*) [Having $\phi_m =1$ would lead to a contradiction: if m terms are (limitwise) equal, adding them up gives a next term m times as large.]

Let $m \in \mathbb{N}$ You can find that

$\frac{F_{n+1}^m}{F_n^m} = \sum_{i=1}^m \frac{F_{n+1-i}^m}{F_n^m} = \sum_{i=0}^{m-1} \frac{F_{n-i}^m}{F_n^m}$

and when you can take the limit

$\lim_{n \rightarrow \infty} \frac{F_{n+1}^m}{F_n^m} = \lim_{n \rightarrow \infty} \sum_{i=0}^{m-1} \frac{F_{n-i}^m}{F_n^m} = \sum_{i=0}^{m-1} \lim_{n \rightarrow \infty} \frac{F_{n-i}^m}{F_n^m} = \sum_{i=0}^{m-1} \phi_m^{-i} = \sum_{i=0}^{m-1} (\frac{1}{\phi_m})^i$ ,

which is a geometric series, so

$\phi_m = \lim_{n \rightarrow \infty} \frac{F_{n+1}^m}{F_n^m} = \frac{1-(\frac{1}{\phi_m})^m}{1-\frac{1}{\phi_m}}$ .

You can then rearrange it to become.

$\phi_m(1-\frac{1}{\phi_m}) = 1-(\frac{1}{\phi_m})^m \Rightarrow \phi_m - 1 = 1-(\frac{1}{\phi_m})^m \Rightarrow \phi_m = 2 + (\frac{1}{\phi_m})^m$

Because $\frac{1}{\phi_m} \geq 0$ it follows that $\phi_m \geq 2$ $\forall m \in \mathbb{N}$

Therefore are $\lim_{m \rightarrow \infty} \phi_m \geq 2$

You can also see that

$\lim_{m \rightarrow \infty} \phi_m = \lim_{m \rightarrow \infty} 2 + (\frac{1}{\phi_m})^m \leq 2 + (\frac{1}{2})^m = 2$

Therefore is

$\lim_{m \rightarrow \infty} \phi_m = 2$

n-nacci sequence has this ratio as the root of x+x^(-n)=2, which is nearest to 2 in the large-n limit. Answer=2.

As shown elsewhere each solution of the m-th sequence can be expressed as phi(m)^m=sum of all lower powers of phi(m).

For m=2, the solution is 1.62 >1.

We can demonstrate that this is an increasing function w.r.t m by multiplying both sides of the equation by phi (m).

Phi (m) ^(m+1) = Sum of lower powers of phi (m) plus 1. Which with a simple derivative allows us to demonstrate that we are dealing with a monotonic increasing function.

We know from the simple solution shown earlier that phi(m)^(m-1)*(phi(m)-2) = -1 and dividing by phi(m)^(m-1) we see that Phi(m)-2 is a very small negative number which approaches 0 at the limit

42 is the answer of the sense of life, so 4 divided by 2 is 2. that was realy easy

include<stdio.h>

float f(int n,int m) {int i,s=0; if(n<m) return 0; if(n==m) return 1; for(i=1;i<=m;i++) { s=s+f(n-i,m);

}

return s;
}

int main() {int m,n; float g;

scanf("%d,%d",&m,&n);

g=f(n+1,m)/f(n,m); printf("%.10f",g); } JUST WRITE THIS C PROGRAMME, AND INPUT SOME VALUE LIKE (60,70) FOR n AND m , you will find the ratio 2