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The sequence given by $a_1 = 3, a_{n+1} = \frac{1}{2}(a_n + 1)$ for all $n \geq 1$. Determine the limit of the sequence as $n \to \infty $. This is only from the first section in the chapter. So, we do not know any advanced theorems on sequences to use yet.

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  • $\begingroup$ Have you heard of the fact that if you can prove that a sequence increases and is bounded above, then it converges, or the fact that if you can prove that a sequence decreases and is bounded below, then it converges? $\endgroup$ – K Split X Apr 30 '17 at 0:34
  • $\begingroup$ Write out a few terms to get an idea. 3, 2, 3/2, 5/4, 9/8,.... look at the denominators. Look at the numerator. That should give a hint. Can prove our hint using induction. Then we can take a guess as to what the limit is L. The we can prove if $n > N$ then $|a_n - L | < \frac 1{2^N}$. $\endgroup$ – fleablood Apr 30 '17 at 0:46
  • $\begingroup$ Well, we have $$\begin{align} &a_1=3\\ &a_2=\frac{1}{2}(a_1+1)=\frac{1}{2}(3+1)=2\\ &a_3=\frac{1}{2}(a_2+1)=\frac{1}{2}(2+1)=\frac{3}{2}=1+\frac{1}{2}\\ &a_4=\frac{1}{2}(a_3+1)=\frac{1}{2}\bigg[\frac{3}{2}+1\bigg]=\frac{5}{4}=1+\frac{1}{4}\\ &a_5=\frac{1}{2}(a_4+1)=\frac{1}{2}\bigg[\frac{5}{4}+1\bigg]=\frac{9}{8}=1+\frac{1}{8}\\ \vdots \end{align}$$ and so the limit is $1$. $\endgroup$ – Juniven Apr 30 '17 at 0:56
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    $\begingroup$ @JohnLou (and whoever else approved it), that edit didn't come out right, so I had to roll it back. $\endgroup$ – dxiv Apr 30 '17 at 1:04
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Hints:

  • $a_2 = \frac{1}{2}(3+1) = 2 \lt 3 = a_1$

  • $\require{cancel} a_{n+1}-a_n = \frac{1}{2}(a_n+\bcancel{1}) - \frac{1}{2}(a_{n-1}+\bcancel{1}) = \frac{1}{2}(a_n - a_{n-1}) = \cdots = \frac{1}{2^{n-1}}(a_2-a_1) \lt 0$

  • $L = \frac{1}{2}(L+1) \iff L = 1$

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Does the sequence increase or decrease (or neither)? Is it bounded above or below (or neither)?

If a sequence increases to an upper bound, it converges. If a sequence decreases to a lower bound, it converges. If a sequence is unbounded or not monotonic (neither increasing nor decreasing), then it diverges.

This is essentially the monotonic sequence theorem.

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Hint: Write a few terms out and so what you get:

$a_1 = 3$

$a_2 = \frac 12(3+1) = 2$

$a_3 = \frac 12(2+1) = \frac 32 = 1 \frac 12$

$a_4 = \frac 12 (1\frac 12 + 1) = \frac 54 = 1 \frac 14$.

So 1) Use induction to prove that $a_n = 1 + \frac 1{2^{n-2}}$

Then 2) Prove for any $0 < \frac 1{2^N} < \epsilon$ then $|a_{N+2} - 1| < \frac 1{2^N} < \epsilon$.

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It is easy to guess that the limit if it exists is $1$ hence it is better to consider the sequence $b_{n}=a_{n}-1$ and then the given relation reduces to $b_{n+1}=b_{n}/2=b_{1}/2^{n}=1/2^{n-1}$ so that $b_{n} \to 0$ and $a_{n} \to 1$. Note that we have also found here an explicit expression for $a_{n} $ as $1+(1/2^{n-2})$ for all $n\geq 1$.

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Rearrange the difference equation to: $a_{n+1}-1=\frac{1}{2}(a_n-1).$

Change $a_n-1=b_n$ to obtain $b_{n+1}=\frac{1}{2}b_n$, which is a geometric progression (GP) with $b_1=a_1-1=3-1=2$ and $q=\frac{1}{2}.$

The $n$-term formula of the GP is $b_n=b_1\cdot q^{n-1}=2\cdot \left(\frac{1}{2}\right)^{n-1}=\left(\frac{1}{2}\right)^{n-2}.$

Refer to the change above to get $a_n=b_n+1=\left(\frac{1}{2}\right)^{n-2}+1.$

Finally, $\lim_\limits{n\to\infty} a_n=\lim_\limits{n\to\infty} \left(\left(\frac{1}{2}\right)^{n-2}+1\right)=1.$

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It's easy to show each $a_n>1.$ Thus $a_{n+1} = (a_n+1)/2 <(a_n+a_n)/2 = a_n.$ Therefore $(a_n)$ is a decreasing sequence that is bounded below. Hence $(a_n)$ converges to a limit $L.$ That $L=1$ falls right out of the familiar end game: $L=(L+1)/2.$

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