Infinity infinite

Calculus Level 4

$^1a_n:1,1,2,3,5,8...$ ,the ratio of two adjacent terms goes to $1.618...(=b_1)$

$^2a_n:1,1,1,3,5,9,17...$ ,the ratio of two adjacent terms goes to $1.839...(=b_2)$

$^3a_n:1,1,1,1,4,7,13,25...$ ,the ratio of two adjacent terms goes to $1.927...(=b_3)$

Each term in the sequence ${}^ka_n$ is obtained by adding together the previous $k+1$ terms in that sequence, with the first $k+1$ terms of that sequence all being equal to $1$

The ratio of two 'infinite adjacent terms' of these sequences forms another sequence $b_k= \lim_{n\to\infty}\frac{^ka_n}{^ka_{n-1} }$ What is $\lim_{n\to\infty}b_n?$

Try my set here .

The answer is 2.0.

1 solution

C Anshul
Jul 17, 2018

$Hints:$

Given $a_{k+2}=a_{k+1}+a_{k}+..........+a_{1}$

The ratio $b=\displaystyle\lim_{n\rightarrow oo}\frac{a_{n}}{a_{n-1}}$

Polynomial in $b$ is

$b^{n}=b^{n-1}+b^{n-2}+..........+b+1$

$b^{n}=\frac{b^{n}-1}{b-1}$

$b^{n+1}-b^{n}=b^{n}-1$

$b=2-b^{-n}$

Applying limit n tends to infty gives

$b=2$

Infinity infinite

Try my set here .

The answer is 2.0.

1 solution

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