Finding $\lim_\limits{n\to\infty}a_n$, using elementary approach without the concept of recurrence, if $a_{n+2}=\frac{a_{n+1}+a_n}2$

Question

The following question exists here but I am looking for a solution using elementary approach without the concept of recurrence.

If $a_1=a,a_2=b,a_{n+2}=\frac{a_{n+1}+a_n}2,n≥0$ then $\lim_\limits{n\to\infty}a_n=\frac{a+\lambda_1b}{\lambda_2};(\lambda_1,\lambda_2\in\mathbb I),$ where $\lambda_1+\lambda_2=?$

Attempt$1$:

$a_3=\frac{a_2+a_1}2=\frac{a+b}2$

$a_4=\frac{a_3+a_2}2=\frac{\frac{a+b}2+b}2=\frac{a+3b}4$

$a_5=\frac{a_3+a_4}2=\frac{\frac{a+b}2+\frac{a+3b}4}2=\frac{3a+5b}8$

$a_6=\frac{5a+11b}{16}$

Not able to write general form of $a_n$ here.

Attempt$2:$

$a_2=a+(b-a)$

$a_3=a+\frac{b-a}2$

$a_4=a+\frac{b-a}2+\frac{b-a}4$

$a_5=a+\frac{b-a}2+\frac{b-a}8$

$a_6=a+\frac{b-a}2+\frac{b-a}8+\frac{b-a}{16}$

Therefore, $a_n$ can't be written like below the way it's mentioned in the linked post.$$a_n=a+\sum_{k=1}^{n}\frac{b-a}{2^k}$$

Attempt$3:$

$a_2=a+(b-a)$

$a_3=a+\frac{b-a}2$

$a_4=a+\frac34(b-a)$

$a_5=a+\frac58(b-a)$

$a_6=a+\frac{11}{16}(b-a)$

Therefore, $a_{n+2}=a+(1-\frac{(2n-3)}{2^n})(b-a)$

Therefore $\lim_\limits{n\to\infty}a_n=a$

But as per the link above, $\lim_\limits{n\to\infty}a_n=a+\frac23(b-a)$

What's wrong in my approach?

Also, as per the above link,

$a_n=(\frac{a+b}3)(1-(-0.5)^{n-2})+\frac b3(1-(-0.5)^{n-3})$

Answer 1

Observe that $$ a_{n+2}-a_{n+1}={-1\over 2}(a_{n+1}-a_n)$$ Therefore $$a_{n+2}-a_{n+1}={(-1)^{n}\over 2^{n}}(a_2-a_1)={(-1)^n\over 2^n}(b-a)$$ Hence $$ a_{n+2}=\sum_{k=0}^{n-1}[a_{k+2}-a_{k+1}]+a_1 =(b-a)\sum_{k=0}^{n-1}{(-1)^k\over 2^k}+a$$ Thus the limits is equal $$(b-a){1\over 1+{1\over 2}}+a={2\over 3}b+{1\over 3}a$$

Answer 2

What is wrong in approach 3 is that you have just guessed a formula of $1-\frac{2n-3}{2^n}$ which fits what you have so far. There's no reason why this should continue to work, and indeed it doesn't: the next one will be $\frac{21}{32}$, not $\frac{25}{32}$ as your formula would suggest.

The correct formula would be $\frac{2^{n+1}+(-1)^n}{3\times 2^n}$, and once you know this you can prove it by induction. But finding the right formula can be difficult without either knowing something about the theory of linear recurrence relations or working out a lot more terms to pin down the pattern.

Answer 3

You can probe by induction that: $a_n=\frac{a+2b}{3}+\frac{4}{3}(b-a)(-\frac{1}{2})^{n}$ therefore the limit is $\frac{a+2b}{3}$.

Also you can get to this formula by: $$2a_3=a_2+a_1$$$$2a_4=a_3+a_2$$$$2a_5=a_4+a_3$$$$2a_{n+1}=a_n+a_{n-1}$$ Adding this equations: $$2a_{n+1}+2a_n=a_1+2a_2+a_n$$ Set $b_n=a_n-\frac{a_1+2a_2}{3}$ => $a_n = b_n + \frac{a_1+2a_2}{3}$ you get $$b_{n+1}=-\frac{b_n}{2}$$ $$b_n=({-\frac{1}{2}})^{n-1}b_1 = ({-\frac{1}{2}})^{n-1}\frac{2}{3}(a-b)=\frac{4}{3}(b-a)(-\frac{1}{2})^{n}$$ Therefore: $$a_n=\frac{a+2b}{3}+\frac{4}{3}(b-a)(-\frac{1}{2})^{n}$$

Answer 4

First, the equation

$$ a_n = a + \sum_{k=1}^n \frac{b-a}{2^k} $$

is incorrect, as you noticed. That was an error by the author of the earlier question.

But your equation

$$ a_{n+2} = a + \left(1 - \frac{2n-3}{2^n} \right) (b-a) $$

is also incorrect.

After guessing a general form from a pattern, check it against the original problem, to see if we can prove it's really true for all terms of the sequence or just so far.

For the earlier question's claim, $a_{n+2} = \frac{a_{n+1}+a_n}{2}$ becomes

$$ a + \sum_{k=1}^{n+2} \frac{b-a}{2^k} \stackrel{?}{=} \frac 12 \left[a + \sum_{k=1}^{n+1} \frac{b-a}{2^k} + a + \sum_{k=1}^n \frac{b-a}{2^k} \right]$$

$$ \frac{b-a}{2^{k+1}} + \frac{b-a}{2^{k+2}} + \sum_{k=1}^n \frac{b-a}{2^k} \stackrel{?}{=} \frac{b-a}{2 \cdot 2^{k+1}} + \sum_{k=1}^n \frac{b-a}{2^k} $$

which is not true (for any $n$). With your claim, $a_{n+2} = \frac{a_{n+1} + a_n}{2} $ becomes

$$ a + \left(1-\frac{2n-3}{2^n}\right)(b-a) \stackrel ?= \frac 12 \left[a + \left(1-\frac{2n-5}{2^{n-1}}\right)(b-a) + a + \left(1-\frac{2n-7}{2^{n-2}}\right)(b-a) \right] $$

$$ -\frac{2n-3}{2^n}(b-a) \stackrel ?= -\frac{2n-5}{2^n}(b-a) - \frac{2n-7}{2^{n-1}}(b-a) $$

which is only true at $n=2$.

When trying to figure out a pattern of a recurrence relation sequence, it's often helpful to resist simplifying the numbers. So in a way between your "attempt 2" and "attempt 3":

$ a_2=a+(b-a) $

$ a_3=a+ \frac 12 (b-a) $

$ a_4=a+ \left(\frac 12 + \frac 14\right)(b-a) $

$ a_5=a+ \left(\frac 12 + \frac 18\right)(b-a) $

$ a_6=a+ \left(\frac 12 + \frac 18 + \frac{1}{16}\right)(b-a) $

$ a_7=a+ \left(\frac 12 + \frac 18 + \frac{1}{32}\right)(b-a) $

$ a_8=a+ \left(\frac 12 + \frac 18 + \frac{1}{32} + \frac{1}{64}\right)(b-a) $

$ a_9=a+ \left(\frac 12 + \frac 18 + \frac{1}{32} + \frac{1}{128}\right)(b-a) $

By now (or maybe not so many are necessary), we can see two patterns showing up, one for odd $n$ and an apparently different one for even $n$:

$$ a_{2m+1} = a + (b-a) \sum_{k=1}^m \frac{1}{2^{2k-1}} $$ $$ a_{2m} = a + (b-a) \sum_{k=1}^{m-1} \frac{1}{2^{2k-1}} + \frac{b-a}{2^{2m-2}} = a_{2m-1} + \frac{b-a}{2^{2m-2}} $$

The sums are geometric series:

$$ a_{2m+1} = a + \frac{\frac 12 - \frac{1}{2^{2m+1}}}{1-\frac 14}(b-a) = \frac 13 a + \frac 23 b - \frac{b-a}{3 \cdot 2^{2m-1}} $$ $$ a_{2m} = \frac 13 a + \frac 23 b - \frac{b-a}{3 \cdot 2^{2m-3}} + \frac{b-a}{2^{2m-2}} = \frac 13 a + \frac 23 b + \frac{b-a}{3 \cdot 2^{2m-2}} $$

Using a $(-1)^n$ to deal with the flipping sign, we can combine both cases into:

$$ a_n = \frac 13 a + \frac 23 b + \frac{b-a}{3} \left(-\frac 12\right)^{n-2} $$

Now to check the answer, $a_{n+2} = \frac{a_{n+1}+a_n}{2}$ becomes

$$ \frac 13 a + \frac 23 b + \frac{b-a}{3} \left(-\frac 12 \right)^n \stackrel ?= \frac 12 \left[\frac 13 a + \frac 23 b + \frac{b-a}{3} \left(-\frac 12 \right)^{n-1} + \frac 13 a + \frac 23 b + \frac{b-a}{3} \left(-\frac 12 \right)^{n-2}\right] $$ $$ \frac{b-a}{3} \left(-\frac 12\right)^n \stackrel ?= -\frac{b-a}{3} \left(-\frac 12\right)^n - \frac{b-a}{3}\left(-\frac 12\right)^{n-1} $$ which is true for all $n$. The equation also holds for $n=1$ ($a_1=a$) and $n=2$ ($a_2=b$), so by induction it is true for all positive integers $n$.

And finally, it's then obvious that $$\lim_{n \to \infty} a_n = \frac13 a + \frac 23 b$$

Answer 5

Especially Lime pointed out my error. Therefore, I reattempted and got the following:

$a_{n+2}=a+\left(1-\frac{2^{n-1}-2^{n-2}+2^{n-3}-...}{2^n}\right)(b-a)$

Taking two cases, odd and even, applying geometric series and then limit, I also get what everyone else has obtained:

$$\frac{a+2b}3$$