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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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marked as duplicate by Arnaud D., gebruiker, J.-E. Pin, trancelocation, Community Aug 1 '18 at 13:00

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  • $\begingroup$ It's determined by $a_1$ and $a_2$ isn't it? Note if $a_1=a_2$ then the sequence is constant. $\endgroup$ – saulspatz Aug 1 '18 at 12:30
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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.

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  • $\begingroup$ Sir i am not expert in linear algebra. To find E we need to find eigen vectors of 1, -1/2 and place then as columns in E right? pls correct me if i am wrong. $\endgroup$ – Magneto Aug 1 '18 at 12:56
  • $\begingroup$ Yes, that's exactly how to do it. :) $\endgroup$ – Dirk Aug 1 '18 at 12:58
  • $\begingroup$ Tq............... $\endgroup$ – Magneto Aug 1 '18 at 13:00
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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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  • $\begingroup$ Sir how did u arrive at that equation? $\endgroup$ – Magneto Aug 1 '18 at 12:59
  • $\begingroup$ Solve the equation $$q^n=\frac{q^{n-1}+q^{n-2}}{2}$$ for $q$ $\endgroup$ – Dr. Sonnhard Graubner Aug 1 '18 at 13:02
  • $\begingroup$ Instead if i add terms using a_n = arithmetic mean i am directly getting limit value. But in ur explanation, how to proceed? i am getting recursively in to powers. It is becoming messy... $\endgroup$ – Magneto Aug 1 '18 at 13:07
  • $\begingroup$ Multiply your equation by $$q^{2-n}$$ $\endgroup$ – Dr. Sonnhard Graubner Aug 1 '18 at 13:09
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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 $2(a_2-a_1)/3+a_1$.

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  • $\begingroup$ can u pls explain me following: i understood how u got 2/3 from given sum. But substract and division part i did not understand. Let $a_1, a_2, a_3, a_4 .....$ be sequence. Then substraction by $a_1$ and division by $a_2$ the sequence becomes $0, (a_2-a_1)/a_2, (a_3-a_1)/a_2 ....$ = $0, 1-(a_1/a_2), ....$ ... Kindly elaborate $\endgroup$ – Magneto Aug 1 '18 at 12:50
  • $\begingroup$ I don’t think you need $a_1\le a_2$. $\endgroup$ – TonyK Aug 1 '18 at 12:50
  • $\begingroup$ @TonyK That is what I said. $\endgroup$ – whatsgoingon Aug 1 '18 at 12:51
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    $\begingroup$ @Magneto From $a_n=\frac{a_{n-1}+a_{n-2}}{2}$ you get that $\frac{a_n-a_1}{a_2-a_1}=\frac{\frac{a_{n-1}-a_1}{a_2-a_1}+\frac{a_{n-1}-a_1}{a_2-a_1}}{2}$. Therefore, the new sequence $b_n=\frac{a_n-a_1}{a_2-a_1}$ also satisfies $b_n=\frac{b_{n-1}+b_{n-2}}{2}$ while $b_1=0$ and $b_2=1$. $\endgroup$ – whatsgoingon Aug 1 '18 at 12:59
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    $\begingroup$ @Magneto Bijective is not enough. You need $F$ to be at least continuous at the limit. Notice however, that the transformation that we used here is affine, and therefore, continuous. But more important, it commutes with the linear difference operator of the differece equation (the recursion). That is why the new sequence satisfies the same equation. $\endgroup$ – whatsgoingon Aug 1 '18 at 13:07

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