general term of sequence.. work shown.. need help!

johnq2k7

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Feb 10, 2009
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A bored student enters the number 0.5 in her calculator, and then repeatedly computes the square of the number in the display. Taking a_0=0.5, find a formula for the general term of the sequence {a_n] of the numbers that appear in the display, and find the limit of the sequence {a_n}

work shown:

the series is a geometric series with a=1/2 and r=1/2

therefore the sigma notation from (n=1 to infinity) of 1/(2^n) = 1

therefore the sum is equal to 1...

and the limit of 1/(2^n) as n goes from 1 to inf. is equal to zero

I'm not sure if my approach and work is correct here please help me with this equation
 
Hello, johnq2k7!

A bored student enters the number 0.5 in her calculator,
and then repeatedly computes the square of the number in the display.
Taking \(\displaystyle a_0 = \tfrac{1}{2}\), find a formula for the general term of the sequence \(\displaystyle \{a_n\}\)
Find the limit of this sequence.

Work shown:

The series is a geometric series with \(\displaystyle a=\tfrac{1}{2}\text{ and }r=\tfrac{1}{2}\) . . . . no

Each term is the square of the previous term: .\(\displaystyle a_n \:=\:\left(a_{n-1}\right)^2\)

\(\displaystyle \text{We have: }\;\begin{array}{ccccc}a_0 &=&\frac{1}{2} \\ a_1 &=&\left(\frac{1}{2}\right)^2 &=& \frac{1}{4} \\ a_2 &=& \left(\frac{1}{4}\right)^2 &=& \frac{1}{16} \\ a_3 &=&\left(\frac{1}{16}\right)^2 &=& \frac{1}{256} \\ \vdots & & \vdots & & \vdots \end{array}\)

\(\displaystyle \text{The }n^{th}\text{ term is: }\:a_n \;=\;\frac{1}{2^{(2^n)}} }\)

\(\displaystyle \text{And: }\:\lim_{n\to\infty}\frac{1}{2^{(2^n)}}}\;=\;\frac{1}{\infty} \;=\;0\)

 
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