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Let the following recursively defined sequence:

$a_{n+1}=\frac{1}{2} a_n +2,$


Prove that $a_n$ converges to 4 by subtracting 4 from both sides.

When I do that, I get: $2(\frac{1}{2} a_{n+1} -2)=(\frac{1}{2} a_n -2)$, so $y=2y$, which is true only for $0$. But I'm not sure how to formally use this in a definition of convergence?

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take a look [… ] – janmarqz Jan 19 '14 at 19:11
i upvoted +1 you all answers, now upvote me! :) – janmarqz Jan 19 '14 at 19:51

4 Answers 4

up vote 1 down vote accepted

Set $b_n=a_n-4$. Then $b_1=-7/2$ and $$ b_{n+1}=a_{n+1}-4=\frac{1}{2}a_n+2-4=\frac{1}{2}(a_n-4)=\frac{1}{2}b_n. $$ Thus $$ b_n=\frac{b_{n-1}}{2}=\frac{b_{n-2}}{2^2}=\cdots=\frac{b_{1}}{2^{n-1}}=-\frac{7}{2^n}, $$ and finally $$ a_n=4-\frac{7}{2^n}. $$ Hence $$ \lim_{n\to\infty} a_n=4. $$

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Just a generalization of the approach for you. Note that $$ a_n=\frac{1}{2}a_{n-1}+2 $$ So if you subtract this expression from the $a_{n+1}$ you have above, you get rid of the constant. Also denote $\Delta a_{n+1}=a_{n+1}-a_{n}$ and you get:

$$ \Delta a_{n+1}=\frac{1}{2} \Delta a_{n}=\frac{1}{2^2} \Delta a_{n-1}=\ldots =\frac{1}{2^{n-1}} \Delta a_{2} $$

If you sum over $n$ the LHS you get a telescoping sum: $\sum_{n=1}^{N}a_{n+1}=a_{N+1}-a_1$. Since $a_2=2 \frac{1}{4}$ and $a_1 = 0.5$ you get (using geometric sum $\sum_{n=1}^{N} \frac{1}{2^{n-1}}=2(1-(\frac{1}{2})^{N+1})$

$$ a_{N+1}=0.5+(2.25-0.5) \cdot 2 \cdot \Bigg(1-(\frac{1}{2})^{N+1} \bigg) $$ and if you take the limit as $N \to \infty$ you get $0.5+3.5=4$.

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You have $2(\frac{1}{2}a_{n+1}-2)=\frac{1}{2}a_{n}-2$. So, you have a sequence $y_n$ where $y_{n+1}=\frac{1}{2}y_n$. This is a bounded decreasing sequence, and hence has a limit, say $y$. Then $y$ satisfies $y=\frac{1}{2}y$ so $y=0$

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note that $a_n\to 4$ when $n\to\infty$.. [… ] – janmarqz Jan 19 '14 at 19:16
Yes- that doesn't contradict what I have mentioned. y=lim $ y_n=0$ and $y_n=(\frac{1}{2}a_n-2)$. So limit of $a_n=4$ – voldemort Jan 19 '14 at 19:17
Define $y_n$ as I mentioned in the answer. $y_{n+1}=\frac{1}{2}y_n$ which is obviously decreasing. Very formally, the ratio of $y_{n+1}/y_n <1$ and hence decreasing. – voldemort Jan 19 '14 at 19:22
@burgundy7 any convergent sequence it is also bounded – janmarqz Jan 19 '14 at 19:24
@burgundy7: This is a decreasing sequence of positive terms. Trivially bounded below by $0$ and bounded above by the first term of the sequence. – voldemort Jan 19 '14 at 19:27

The solution to this recurrence is $$a_n=4-\frac{7}{2^n},$$ so $a_n$ converges to $4$.

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