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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.

Magneto
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3 Answers3

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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.

Dirk
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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.

Dr. Sonnhard Graubner
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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 $2(a_2-a_1)/3+a_1$.

whatsgoingon
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  • 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 – Magneto Aug 01 '18 at 12:50
  • I don’t think you need $a_1\le a_2$. – TonyK Aug 01 '18 at 12:50
  • @TonyK That is what I said. – whatsgoingon Aug 01 '18 at 12:51
  • I mean, you don’t need to worry about exchanging their roles. – TonyK Aug 01 '18 at 12:52
  • @Magneto It should be "divide by $a_2-a_1$". – whatsgoingon Aug 01 '18 at 12:52
  • @whatsgoingon Still it does not work. Denominator $a_2-a_1$ will be present through out. Ur's is a specific case of numbers matching the given sequence pattern. Good thought though. Tq. – Magneto Aug 01 '18 at 12:55
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    @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 $b_2=1$. – whatsgoingon Aug 01 '18 at 12:59
  • I got what u r saying.....can u pls tell me 1 thing? If $a_1, a_2, a_3....$ is a sequence, and if F is a bijective function to another sequence $b_1, b_2, b_3 ....$ can we say both domain and range converge to same limit??? Pls explain – Magneto Aug 01 '18 at 13:03
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    @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. – whatsgoingon Aug 01 '18 at 13:07
  • Got it ... Tq sir ... nice thought process.... – Magneto Aug 01 '18 at 13:12