Correction needed: Sequences

Question

I just messed up my Advanced levels Further maths exam(Revising Maths day before the exam isn't quite a good Idea)...I'd like to get the correction of This question(Learn from your mistakes they say)

The following sequences are defined as follows

U₀=1, U_n+1=(U_n+2V_n)/3

V₀=12, V_n+1=(U_n+3V_n)/4

W_n=V_n-U_n

i) Show that W_n is a G.P with Positive terms

ii)Find the limit of the sequence W_n

iii)Show that U_n is increasing

iv)Show that V_n is decreasing

v)Hence Show that U₀≤U_n≤V₀≤V_n

vi)Show that the sequences U_n and V_n converge to L.

Another sequence t_n is defined by t_n=3U_n+8V_n

vii) Show that t_n is a constant sequence and hence find L.

For the (i) part I found an expression for W_n+1 in terms of U_n+1 and V_n but couldn't show It was a GP

For the other question...I know how to prove monotony and Bounds for a recursive sequences...But this one is defined in terms of another recursive sequence

please really need Help.

Answer 1

A direct calculation gives :

$U_n=4V_{n+1}-3V_n$

Reporting in first equation $2V_n+U_n=3U_{n+1}\iff 2V_n+4V_{n+1}-3V_n=12V_{n+2}-9V_{n+1}$

Thus : $12V_{n+2}-13V_{n+1}+V_n=0$

Characteristic equation $12r^2-13r+1=(12r-1)(r-1)$

So $V_n=a+\dfrac b{12^n}$ and $U_n=a-\frac 83\dfrac b{12^n}$

Solving for $U_0=1,V_0=12$ gives $a=9,b=3$ so $U_n=9-\dfrac 8{12^n}$ and $V_n=9+\dfrac 3{12^n}$

And their difference $W_n$ is GP because the constant term disappear : $W_n=V_n-U_n=\dfrac {11}{12^n}$

---

Let's go back to the method that was proposed in the exercise:

i)

$12W_{n+1}=12V_{n+1}-12U_{n+1}=(3U_n+9V_n)-(4U_n+8V_n)=V_n-U_n=W_n$

ii)

Thus $W_n=\dfrac{V_0-U_0}{12^n}=\dfrac{11}{12^n}\to 0$ and $W_n>0$

iii)

$3U_{n+1}-3U_n=U_n+2V_n-3U_n=2(V_n-U_n)=2W_n>0$ so $U_n\nearrow$

iv)

$4V_{n+1}-4V_n=U_n+3V_n-4V_n=U_n-V_n=-W_n<0$ so $V_n\searrow$

v)

There is a mistyping here, should be $U_0\le U_n\le V_n\le V_0$

• first inequality because $U_n$ increasing
• middle inequality because $W_n>0$
• last inequality because $V_n$ decreasing

vi)

$(U_n)_n\nearrow$ and upperly bounded by $V_0$ so the sequence converges.

$(V_n)_n\searrow$ and lowerly bounded by $U_0$ so the sequence converges.

Since their difference $W_n\to 0$ both sequences are adjacent and converge to the same limit $L$.

vii)

$T_{n+1}=3U_{n+1}+8V_{n+1}=(U_n+2V_n)+(2U_n+6V_n)=3U_n+8V_n=T_n$

Thus $T_n\to 3L+8L=11L$ but since constant also to $T_0=3U_0+8V_0=99$ so $L=9$