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I'm having a really hard time finding the limit of a recursive sequence -

$$ \begin{align*} &a(1)=2,\\ &a(2)=5,\\ &a(n+2)=\frac12 \cdot \big(a(n)+a(n+1)\big). \end{align*}$$

I proved that the sequence is made up from a monotonically increasing sequence and a monotonically decreasing sequence, and I proved that the limits of the difference of these sequences is zero, so by Cantor's Lemma the above sequence does converge. I manually found out that it converges to $4$, but I can't seem to find any way to prove it.

Any help would be much appreciated! Thank you.

Chappers
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Adam
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  • related to http://math.stackexchange.com/questions/468385/solving-recurrence-equation-using-generating-function-method – Evan Aug 15 '13 at 17:16
  • Look at $a(n)-4$; you should be able to (experimentally) find a simple equation for these values, and once you have such an equation it's easy to prove it. – Steven Stadnicki Sep 29 '17 at 22:59

4 Answers4

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It never hurts to look at some data. Calculate the first few terms:

$$2,5,\frac72,\frac{17}4,\frac{31}8,\frac{65}{16},\frac{127}{32}$$

We can even fit the first two numbers into the apparent pattern:

$$\frac1{1/2},\frac51,\frac72,\frac{17}4,\frac{31}8,\frac{65}{16},\frac{127}{32}$$

The denominator of $a(n)$ appears to be $2^{n-2}$, and the numerators are alternately one less and one more than a power of $2$. Specifically, it appears that the numerator of $a(n)$ is $2^n-1$ when $n$ is odd and $2^n+1$ when $n$ is even or, more simply, $2^n+(-1)^n$. We conjecture, therefore, that

$$a(n)=\frac{2^n+(-1)^n}{2^{n-2}}\;.$$

Now prove this by induction on $n$, and observe that it’s now trivial to calculate the limit of the sequence.

Brian M. Scott
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We can write the general term as $$\tag1 a(n) = u+\left(-\frac12\right)^n\cdot v$$ (because $1$ and $-\frac12$ are the two solutions of $x^2=\frac12(x+1)$). From $a(1)=2=u-\frac12 v$ and $a(2)=5=u+\frac14 v$, we find $u=v=4$. From (1) it is clear that $$\lim_{n\to\infty} a(n)=u=4$$

Hagen von Eitzen
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First, given $\mathbf{a}_i=[a_i,a_{i+1}]^T$, you can express the sequence as $\mathbf{a}_{i+1}=M\mathbf{a}_i$. Find the infinite power of $M$ by diagonalizing it and raising the diagonal term to an infinite power. Then the limit is $M^\infty\mathbf{a}_0$.

Sneftel
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    I'm trying to imagine someone who is unable to answer the question on his own but who can understand and use this answer. So far, my imagination has not succeeded. – Andreas Blass Aug 15 '13 at 18:54
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giving a little elaboration to Hagen's answer: note that what we have here is a linear recurrence relation and assume a solution of the form $a(n)=x^n$. Observe:$$a(n+2)=\frac12(a(n)+a(n+1))\\x^{n+2}=\frac12(x^n+x^{n+1})\\x^{n+2}-\frac12x^{n+1}-\frac12x^n=0\\x^n\left(x^2-\frac12x-\frac12\right)=0$$... which yields the trivial solution $x=0$ but also the solutions $x=1,-1/2$, i.e. $a(n)=1$ and $a(n)=(-1/2)^n$ are two solutions to the above recurrence.

Because our problem is linear it follows any linear combination of $1,(-1/2)^n$ satisfies our equation, giving the general solution $a(n)=u+v(-1/2)^n$.

Imposing our initial conditions $a(1)=2,a(2)=5$ we determine our coefficients $u=v=4$ giving the solution $a(n)=4+4(-1/2)^n$. Clearly $a(n)\to4$ in the limit as $n\to\infty$.

Community
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obataku
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