Infinite Series Problem (I'm stuck)

[DivGradCurl]

\textrm{(a) A sequence}\left\{ a_n \right\} \textrm{is defined recursively by the equation} a_n = \frac{1}{2} \left( a_{n-1} + a_{n-2} \right) \textrm{for} n \geq 3 \textrm{, where} a_1 \textrm{and} a_2 \textrm{can be any real numbers. Experiment with various values of} a_1 \textrm{and} a_2 \textrm{and use your calculator to guess the limit of the sequence.}

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\textrm{Here is what I've got:}

a_1 = 1 \qquad a_2 = 3 \qquad \Longrightarrow a_n = 2.\overline{3}

a_1 = 7 \qquad a_2 = 16 \qquad \Longrightarrow a_n = 13

a_1 = -3 \qquad a_2 = 48 \qquad \Longrightarrow a_n = 31

a_1 = 0 \qquad a_2 = 1 \qquad \Longrightarrow a_n = 0.\overline{6}

a_1 = 0.3 \qquad a_2 = 0.98 \qquad \Longrightarrow a_n = 0.75\overline{3}

a_1 = 4 \qquad a_2 = 6 \qquad \Longrightarrow a_n = 5.\overline{3}

\textrm{Is that it? I'm not so sure about that.}

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\textrm{(b) Express} \lim _{n \to \infty} a_n \textrm{in terms of} a_1 \textrm{and} a_2 \textrm{by expressing} a_{n-1} - a_n \textrm{in terms of} a_2 - a_1 \textrm{and summing a series.}

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\textrm{Since I've had some difficulty finding a common pattern, I don't even have a clue!}

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\textrm{Thanks!}

 

[maverick280857]

0. Reminds me of Fibonacci...interesting...

1. Can a_{1} = a_{2}? Does your problem permit that?

Lets see...

a_{1} = 1
a_{2} = 1
a_{3} = 0.5(a_{2} + a_{1}) = 1
a_{4} = 0.5(a_{3} + a_{2}) = 1
a_{5} = 0.5(a_{4} + a_{3}) = 1
a_{6} = 0.5(a_{5} + a_{4}) = 1
...
a_{n} = 1!

Of course if a_{1} \neq a_{2} then that's a different story. And it seems it is indeed so for you have not taken the first two terms in your solutions as being equal.

Um...let's see what can fit in...

Cheers
Vivek

 

[Tide]

Subtract

a_n

from both sides of your recurrence relation to give

a_n - a_{n-1} = -\frac {1}{2} (a_{n-1} - a_{n-2})

which you will recognize as a geometric series in the difference of successive elements in your series.

You should be able to see your way through the rest. I ended up with

a_n = \frac {a_1 + 2 a_2}{3} - \left( - \frac {1}{2} \right)^n \frac {4 (a_1 - a_2)}{3}

 

[DivGradCurl]

This is as far as I could get:

a_n = \frac{1}{2} \left( a_{n-1} + a_{n-2} \right)

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If:

a_1 = 1

a_2 = 3 \Longrightarrow a_2 - a_1 = 2

a_3 = 2 \Longrightarrow a_3 - a_2 = -1

a_4 = \frac{5}{2} \Longrightarrow a_4 - a_3 = \frac{1}{2}

a_5 = \frac{9}{4} \Longrightarrow a_5 - a_4 = -\frac{1}{4}

a_6 = \frac{19}{8} \Longrightarrow a_6 - a_5 = \frac{1}{8}

It follows that:

a_n - a_{n-1} = \left( a_2 - a_1 \right) \sum _{n=1} ^{\infty} \left( - \frac{1}{2} \right) ^{n-1} = \frac{2}{3} \left( a_2 - a_1 \right)

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a_n - a_{n-1} = \frac{1}{2} a_{n-1} - a_{n-1} + \frac{1}{2} a_{n-2}

a_n - a_{n-1} = -\frac{1}{2} \left( a_{n-1} - a_{n-2} \right)

\frac{2}{3} \left( a_2 - a_1 \right) = -\frac{1}{2} \left( a_{n-1} - a_{n-2} \right)

\frac{2}{3} \left( a_2 - a_1 \right) = -\frac{1}{2} \left[ \left( a_n - a_{n-2} \right) - a_{n-2} \right]

a_n = - \frac{4}{3} \left( a_2 - a_1 \right)

What additional steps are necessary to find:

a_n = \frac {a_1 + 2 a_2}{3} - \left( - \frac {1}{2} \right)^n \frac {4 (a_1 - a_2)}{3} ? Do I miss the whole point?

 

[Tide]

Starting with

a_n - a_{n-1} = -\frac {1}{2} (a_{n-1} - a_{n-2})

it follows that

a_n = a_{n-1} + \left(-\frac {1}{2}\right)^{n-2} (a_2-a_1)

Now just recurse a_{n-1} down to index 1:

a_n = a_1 + (a_2-a_1) \sum_{j=0}^{n-2} \left( - \frac {1}{2}\right)^j

You can sum the geometric series and rearrange terms to arrive at the result I showed earlier.

 

[DivGradCurl]

I finally get the same result.

Thank you very much.