If a sequence is such that $a_k$ is arithmetic mean of its two immediately preceding terms. Show that the sequence converge. Find the limit of sequence
I got to see that the odd sub sequence is increasing and bounded above by $a_2$ and even sub sequence decreasing and bounded below by $a_1$. Now the sequence is convergent. Bow how to find the limit of sequence.
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Hint: Make the ansatz $$a_n=q^{\lambda}$$ since the equation is linear. For your Control: The solution is given by $$a_n=\left(\frac{-1}{2}\right)^nC_1+C_2$$ where $C_1,C_2$ are constants.
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You can compute the sequence. It looks like you assumed $a_1\leq a_2$. If not, you can exchange their roles. You can subtract $a_1$ from the sequence and divide by $a_2-a_1$ to assume that $a_1=0$ and $a_2=1$. Then, the sequence of points would be $0,1,1/2,3/4,5/8,..., 0+1-1/2+1/4-1/8+1/16-...$ which is a geometric progression of ratio $-1/2$. Its sum would be $1/(1+1/2)=2/3$.
To return to the original sequence you can multiply by $a_2-a_1$ and add $a_1$.
Then the limit is $2(a_2-a_1)/3+a_1$.
As this is a linear recursion, linear algebra can help you here:
We have $$\begin{pmatrix}a_{n} \\ a_{n-1}\end{pmatrix} = \begin{pmatrix}1/2 & 1/2 \\ 1 & 0\end{pmatrix}\begin{pmatrix}a_{n-1} \\ a_{n-2}\end{pmatrix}.$$ So setting $A := \begin{pmatrix}1/2 & 1/2 \\ 1 & 0\end{pmatrix}$, we have that $$\begin{pmatrix}a_{n} \\ a_{n-1}\end{pmatrix} = A^{n-1}\begin{pmatrix}a_2 \\ a_1\end{pmatrix}.$$
Now $A$ has eigenvalues $1$ and $-1/2$, so diagonalizing $A$ we get $A = EDE^{-1}$ with $D = diag(1,-1/2)$ and $E$ some matrix (to be computed). Putting this in, we get $$\begin{pmatrix}a_{n} \\ a_{n-1}\end{pmatrix} = ED^{n-1}E^{-1}\begin{pmatrix}a_2 \\ a_1\end{pmatrix}.$$
Now the powers of a diagonal matrix are easy to compute, so we have a closed, recursion-free formula for $a_n$, once we have computed $E$.
Note that this method works in many cases where the recursive rule is linear.