Convergence of Sequence of arithmetic mean

Question

If $a_{n+2}=\frac {a_{n+1}+a_{n}}{2}$ $\forall n$>0 ,I have to show that $a_{n}\to \frac {a_{1}+2a_{2}}{3}$.


I don't know this problem is easy or difficult as intially I was posting here my query about question but at the time of writting problem I got this solution .If any mistake please tell me ..

Answer 1

First of all to show existence of limit We have term $a_1,a_2$ we know that given seunce is of artihematice mean or we also show that min{$a_1,a_2$} < a3 < max{$a_1,a_2$}
WLOG say $a_1 < a_3 < a_2$
Similarly to above we can say that $a_3 < a_4 < a_2$ And by induction we can easily show that given sequence is monotonic and bounded above so Must be convergent to say a.
$a_{n+2}-a_{n+1}$=$\frac {a_{n}-a_{n+1}}{2}$
Taking Sum for index n we get as this alternating series
$\sum_{i=1}^n a_{i+2}-a_{+1}$=$\sum_{i=1}^n\frac {a_{i}-a_{i+1}}{2}$
So we left with $a_{n+2}-a_{2}$=$\frac {a_{1}-a_{n+1}}{2}$ By Transferring We left with what is needed $a_{n+2}-a_{n+1}/2$=$\frac {a_{1}+2a_{2}}{2}$
As $a_n \to a$ So 3$a_n$/2=$\frac {a_{1}+2a_{2}}{2}$ Which leads to required solution.

Answer 2

Write $b=(a_1+2a_2)/3$ and $c=a_1-b$. Then $a_2-b=(-1/2)c$, $a_3-b=(1/4)c$ and in general $a_n-b=(-1/2)^{n-1}c$. So $a_n\to b$.

Answer 3

The characteristic equation: $2\lambda^2-\lambda -1=0$ has 2 solutions: $1$ and $\frac{-1}{2}.$

Thus, $a_{n}=A+B.\left(\frac{-1}{2}\right)^n,$ where $A$ and $B$ are constants such that $a_1=A-\frac{B}{2}, a_2=A+\frac{B}{4}.$ So, $A=\frac{a_1+2a_2}{3}.$

Moreover, $\{a_n\}$ converges to $A=\frac{a_1+2a_2}{3}.$

Answer 4

If you consider $$a_{n+2}=\frac {a_{n+1}+a_{n}}{2}$$ the characteristic equation is $r^2=\frac{r+1}2$, the roots of which being $r_1=-\frac 12$ and $r_2=1$. So, the genaral solution is $$a_n=c_1 \left(-\frac{1}{2}\right)^n+c_2$$ If we impose $a_1=A$, $a_2=B$, this leads to

$$a_n=\frac 13 \left(A+2B+(-1)^n\frac{(B-A)}{2^{n-2}} \right)$$

Answer 5

Observe that

$$a_{n+1}-a_n=\frac{a_n-a_{n-1}}{-2}$$ and by induction

$$a_{n+1}-a_n=\frac{a_2-a_1}{(-2)^{n-1}}.$$

Then

$$a_{n}-a_1=(a_{n}-a_{n-1})+(a_{n-1}-a_{n-2})+\cdots(a_2-a_1)\\ =(a_2-a_1)\left(1-\frac12+\frac14-\cdots\frac1{(-2)^{n-2}}\right)$$

where the summation tends to $\dfrac23$. You can draw $a_\infty$.

Answer 6

Let $a_1=a_2=1$. The next terms are obviously $1,1,1,\cdots$.

Now let $a_1=2,a_2=-1$. The next terms are $\frac12,-\frac14,\frac18,\cdots \left(-\frac12\right)^k,\cdots$ (because $\frac{\left(-\frac12\right)^0+\left(-\frac12\right)^1}2=\left(-\frac12\right)^2$).

Now for arbitrary $a_1,a_2$, find the linear combination such that

$$p\,(1,1)+q\,(2,-1)=(a_1,a_2).$$

From the linear system, $3p=a_1+2a_2$ and the limit is $1\,p+0\,q$.