MHB What is the limit of xn in this arithmetic mean sequence?

[aruwin]

I have a problem on how to generalize this sequence.
Problem:
In sequence a,b,x1,x2,...xn,...,each term is equals to the arithmetic mean of its two preceding numbers.Using a and b, find:

1.xn

2.lim(n->∞) xn

My working:

x₁ = (a+b)/2, x₂=(b+x₁)/2
x₃ = (x₂+x₁)/2...

Ok, so I did the same thing as above until x₅ and substituted all that are needed to be substituted with a and b.And here's what I get:

(a+b)/2,(a+3b)/4,(3a+5b)/8,(5a+11b)/16,(11a+21b)/32+...I realize a pattern here. Obviously, the denominator is just increasing by 2n. I see a pattern in the numerator too,but it's hard to generalize it using n.I realize that the value of the second a is equals to the value of the previous b and the value of the third a is equals to the second b and it goes on.
You see it too,don't you?When the first term has b in the numerator,then the second term has a, and when the 2nd term has 3b in the numerator, the 3rd term has 3a and so on...now we just have to generalize that,don't we?

 

[Sudharaka]

I have a problem on how to generalize this sequence.
Problem:
In sequence a,b,x1,x2,...xn,...,each term is equals to the arithmetic mean of its two preceding numbers.Using a and b, find:

1.xn

2.lim(n->∞) xn

My working:

x₁ = (a+b)/2, x₂=(b+x₁)/2
x₃ = (x₂+x₁)/2...

Ok, so I did the same thing as above until x₅ and substituted all that are needed to be substituted with a and b.And here's what I get:

(a+b)/2,(a+3b)/4,(3a+5b)/8,(5a+11b)/16,(11a+21b)/32+...I realize a pattern here. Obviously, the denominator is just increasing by 2n. I see a pattern in the numerator too,but it's hard to generalize it using n.I realize that the value of the second a is equals to the value of the previous b and the value of the third a is equals to the second b and it goes on.
You see it too,don't you?When the first term has b in the numerator,then the second term has a, and when the 2nd term has 3b in the numerator, the 3rd term has 3a and so on...now we just have to generalize that,don't we?

Hi arwin, :)

I am not sure whether you will be able to write the general pattern of the sequence through looking at the first few terms. However, the general formula could be found out by solving a recurrence relation. The nth term of the sequence is given by the recurrence relation,

\[x_n=\frac{x_{n-1}+x_{n-2}}{2}\]

This is a recurrence relation of the second order and could be solved by the substituting, \(x_{n}=r^{n}\mbox{ where }n>2\). Then, the characteristic polynomial would be,

\[r^n=\frac{r^{n-1}+r^{n-2}}{2}\]

\[\Rightarrow 2r^2=r+1\]

\[\Rightarrow r=1,\,-\frac{1}{2}\]

Therefore the general solution would be,

\[x_n=A+B\left(-\frac{1}{2}\right)^n\mbox{ where A and B are constants.}\]

Using, \(\displaystyle x_3=\frac{3a+5b}{8},\,x_{4}=\frac{5a+11b}{16}\) you can get,

\[A=\frac{a+2b}{3}\mbox{ and }B=\frac{b-a}{3}\]

\[\therefore x_n=\frac{a+2b}{3}+\frac{b-a}{3}\left(-\frac{1}{2}\right)^n\mbox{ where }n>2\]

By substituting \(n=1\mbox{ and }2\) in the above equation we get, \(\displaystyle x_{1}=\frac{a+b}{2}\mbox{ and }x_{2}=\frac{a+3b}{4}\) which is the first term and the second term of the sequence. Therefore the formula is also valid for \(n=1\mbox{ and }n=2\).

\[\therefore x_n=\frac{a+2b}{3}+\frac{b-a}{3}\left(-\frac{1}{2}\right)^n\mbox{ where }n\in\mathbb{Z}^+\]

Kind Regards,
Sudharaka.

 

[aruwin]

Thanks!You're so smart! I got it now :) Just one more thing, what about the limit? Since A+B^n, looks like it has no limit because it becomes infinity. What to do?

 

[Sudharaka]

Thanks!You're so smart! I got it now :) Just one more thing, what about the limit? Since A+B^n, looks like it has no limit because it becomes infinity. What to do?

You are welcome. :)

\[x_n=\frac{a+2b}{3}+\frac{b-a}{3}\left(-\frac{1}{2}\right)^n\]

\[\Rightarrow\lim_{n\rightarrow\infty}x_{n}=\frac{a+2b}{3}+\frac{b-a}{3}\lim_{n\rightarrow\infty}\left(-\frac{1}{2}\right)^n\]

So do you know what happens to \(\displaystyle\left(-\frac{1}{2}\right)^n\mbox{ when }n\rightarrow\infty\) ?

 

[aruwin]

You are welcome. :)

\[x_n=\frac{a+2b}{3}+\frac{b-a}{3}\left(-\frac{1}{2}\right)^n\]

\[\Rightarrow\lim_{n\rightarrow\infty}x_{n}=\frac{a+2b}{3}+\frac{b-a}{3}\lim_{n\rightarrow\infty}\left(-\frac{1}{2}\right)^n\]

So do you know what happens to \(\displaystyle\left(-\frac{1}{2}\right)^n\mbox{ when }n\rightarrow\infty\) ?

OK,I got it!Since the numbers turn out to be positive and negative alternately, it would be 0! So only the left side is left, and the limit will be (a+2b)/3. Correct?

 

[Sudharaka]

OK,I got it!Since the numbers turn out to be positive and negative alternately, it would be 0! So only the left side is left, and the limit will be (a+2b)/3. Correct?

Indeed. \(\left(-\frac{1}{2}\right)^n\) is an alternating sequence which converges to zero.

\[\therefore\lim_{n\rightarrow\infty}x_{n}=\frac{a+2b}{3}\]