[College Calc 01-02. Limits of Sequences] #09

Let $d$ be the distance between the surfaces; the time it takes for the particle to travel back and forth with average speed $a_n$ and $a_{n+1}$ respectively is

$t = \frac{d}{a_n} + \frac{d}{a_{n+1}}.$

Then, the average speed of the particle is

$\frac{2d}{t} = \frac{2d}{\frac{d}{a_n} + \frac{d}{a_{n+1}}} = \frac{2}{\frac{1}{a_n} + \frac{1}{a_{n+1}}};$

the harmonic mean of $a_n$ and $a_{n+1}.$

Note that the average speed during the first two travels equals $\frac{2}{\frac{1}{120}+\frac{1}{40}} = 60(\rm{m/s}).$ Since this needs to half down every time $n$ increases by $1,$ we may write this as

$\frac{2}{\frac{1}{a_n}+\frac{1}{a_{n+1}}} = 60\cdot \left(\frac{1}{2}\right)^{n-1},$

or simply,

$\frac{1}{a_n}+\frac{1}{a_{n+1}} = \frac{2^n}{60}\quad\cdots\quad \rm{Eq1}(n).$

Let us now write $b_n = \frac{1}{a_n}$ for readability. Note that subtracting $\rm{Eq1}(n-2)$ from $\rm{Eq1}(n-1)$ yields

$\begin{aligned}b_n - b_{n-2} &= \frac{2^{n-1}-2^{n-2}}{60} \\ &= \frac{2^{n-2}}{60}\quad\cdots\quad\rm{Eq2}(n)\end{aligned}$

For $n=2k,$ adding the equations $\rm{Eq2}(4),\rm{Eq2}(6),~\cdots,~\rm{Eq2}(n)$ yields

$b_{2k} = b_2 + \frac{2^2}{60}\frac{4^{k-1} - 1}{3} = \frac{2^{2k+1}+1}{360}.$

Similarly, we have

$b_{2k-1} = b_1 + \frac{2^1}{60}\frac{4^{k-1}-1}{3} = \frac{2^{2k}-1}{360}.$

In general,

$b_n = \frac{2^{n+1} + (-1)^n}{360}.$

(This can also be proved by finding some appropriate $a$ such that $c_n + c_{n+1} = (b_n + a\cdot 2^n) + (b_{n+1} + a\cdot 2^{n+1}) = 0;$ such a solution is left as an exercise for the readers.)

Now for the desired limit, we first evaluate

$\begin{aligned} a_n - 2a_{n+1} &= \frac{360}{2^{n+1} + (-1)^n} - \frac{720}{2^{n+2} + (-1)^{n+1}} \\ &= 360\cdot\frac{(-1)^{n+1} - 2(-1)^n}{(2^{n+1} + (-1)^n)(2^{n+2} + (-1)^{n+1})} \\ &= \frac{-1080(-1)^n}{(2^{n+1} + (-1)^n)(2^{n+2} + (-1)^{n+1})}. \end{aligned}$

Then, we may easily have

$\begin{aligned} \lim_{n\to\infty} 4^n|a_n - 2a_{n+1}| &= \lim_{n\to\infty} 4^n \left|\frac{-1080(-1)^n}{(2^{n+1} + (-1)^n)(2^{n+2} + (-1)^{n+1})}\right| \\ &= \lim_{n\to\infty} 4^n \frac{1080}{2^{2n+3}} \\ &= \boxed{135}. \end{aligned}$

If the distance between the surfaces is $s$ and the average speed from trip $n-1$ and trip $n$ is $v_n$ , then we have: $v_n=\frac{2s}{t_{n-1}+t_n}=\frac{2s}{\frac{s}{a_{n-1}}+\frac{s}{a_n}}=\frac{2a_{n-1}a_n}{a_{n-1}+a_n} \ \ \ (1)$ We are given that $2v_{n+1}=v_n$ . Substituting $(1)$ in this equation we get: $\frac{4a_na_{n+1}}{a_n+a_{n+1}} =\frac{2a_{n-1}a_n}{a_{n-1}+a_n}$ $a_{n-1}(a_n+a_{n+1}) =2a_{n+1}(a_{n-1}+a_n)$ $a_{n+1} =\frac{a_na_{n-1}}{2a_n+a_{n-1}}$ Upon dividing both the numerator and the denominator by $a_na_{n-1}$ and taking the reciprocal, we get: $\frac{1}{a_{n+1}}=\frac{1}{a_n}+\frac{2}{a_{n-1}}$ For simplicity we denote $b_n=\frac{1}{a_n}$ and the above equation is rewritten in: $b_{n+1} =b_n+2b_{n-1}\Leftrightarrow b_{n+2}=b_{n+1}+2b_n\ \ \ (2)$ $b_1 =\frac{1}{a_1}=\frac{1}{120}$ $b_2 =\frac{1}{a_2}=\frac{1}{40}$ This looks very similar to the Fibonacci sequence. We'll derive $b(n)$ in a way, similar to how Binet's formula for Fibonacci numbers is derived. First, suppose $b_n$ is a geometric progression with a quotient $k$ . We can make this supposition, because calculating the first 10-15 elements of $b_n$ we notice that the quotient $\frac{b_{n+1}}{b_n}$ converges. In other words as $n\to\infty\ b_n$ turns in a geometric progression. So let's use $b_n=b_1k^{n-1}$ and substitute this in $(2)$ : $b_1k^{n+1}=b_1k^n+2b_1k^{n-1} |:b_1k^{n-1}\neq 0$ $k^2-k-2=0$ This quadratic has 2 solutions: $k=2; k=-1$ , which explains why $b_n$ isn't really a geometric progression. Now we write $b_n=A.2^n+B(-1)^n$ . To find the values of the coefficients $A$ and $B$ we'll use what we found for $b_1$ and $b_2$ : $\begin{cases} b_1=A.2+B(-1)=\frac{1}{120} \\ b_2=A.2^2+B(-1)^2=\frac{1}{40} \end{cases}$ From this system we find $A=\frac{1}{180}, B=\frac{1}{360}$ and so: $b_n=\frac{2^n}{180}+\frac{(-1)^n}{360}=\frac{2.2^n+(-1)^n}{360}=\frac{2^{n+1}+(-1)^n}{360}$ $\Rightarrow a_n=\frac{1}{b_n}=\frac{360}{2^{n+1}+(-1)^n}$ Let $X=a_n-2a_{n+1}$ . $X=\frac{360}{2^{n+1}+(-1)^n}-\frac{720}{2^{n+2}+(-1)(-1)^n}$ $X=\frac{360(2^{n+2}-(-1)^n-2.2^{n+1}-2(-1)^n)}{(2.2^n+(-1)^n)(4.2^n-(-1)^n)}$ $X=\frac{360(2^{n+2}-2^{n+2}-3(-1)^n)}{2^n.2^n(2+(-\frac{1}{2})^n)(4-(-\frac{1}{2})^n)}$ $X=\frac{-1080(-1)^n}{4^n(2+(-\frac{1}{2})^n)(4-(-\frac{1}{2})^n)}$ Taking the absolute value of the expression above results in the removal of all minus signs in the numerator and transforming $(-1)^n$ into $1$ or: $|X|=\frac{1080}{4^n(2+(-\frac{1}{2})^n)(4-(-\frac{1}{2})^n)}$ $\lim_{x\to\infty} 4^n|X|=\lim_{x\to\infty} \frac{4^n1080}{4^n(2+(-\frac{1}{2})^n)(4-(-\frac{1}{2})^n)}=\frac{1080}{(2+0)(4-0)}=\fbox{135}$ In the limit we used that for $|x|<1$ $\lim_{n\to\infty} x^n=0$ .

Set our time unit such that the first traversal takes one unit of time. We get $t_1=1, t_2=3, t_3=5$ etc.

The halving velocity criterion can be expressed as a double time criterion: $t_{n+1}+t_{n}=2(t_{n}+t_{n-1})$

From this we get $t_{n+1} = 2t_{n-1} + t_{n}$

It it easy to show that $t_n = 2t_{n-1}+1 \Rightarrow t_{n+1} =  2t_{n}-1$ and similarly $t_n = 2t_{n-1}-1 \Rightarrow t_{n+1} = 2t_{n}+1$

From the initial values and the recurrence relations, we conclude that $t_n$ are the Jacobsthal numbers : $t_{n-1} = J_n = \frac{2^n-(-1)^n}{3}$

We can now rewrite the expression

$a_n-2a_{n+1} = 120(\frac{3}{2^{n+1}-(-1)^{n+1}}-2\frac{3}{2^{n+2}-(-1)^{n+2}})$ $= 360\frac{2^{n+2}-(-1)^{n+2} - 2(2^{n+1}-(-1)^{n+1})} {(2^{n+1}-(-1)^{n+1})(2^{n+2}-(-1)^{n+2})}$ $= 360\frac{3(-1)^{n+1}} {(2^{2n+3}-2^{n+2}(-1)^{n+1})-(-1)^{n+2}2^{n+1}+(-1)^{2n+3}}$

In the limit, only the highest order powers remain, so $\lim_{n \to \infty} 4^n|a_{n}-2a_{n+1}|=\lim_{n \to \infty} 2^{2n}×360×\frac{3}{2^{2n+3}}=360×\frac{3}{8}=\boxed{135}$