What is the series an
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Dr.Peterson
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I presume what you want to find is the sum, presumably as an explicitly defined function of x.
Can you tell us what you have been learning, including what particular methods you have seen?
First of all, the definition of [imath]a_n[/imath] is slightly ambiguous.
[math]a_n = \dfrac{\dfrac{1}{2} * \displaystyle \prod_{j=1}^n \left ( \dfrac{1}{2} - j \right )}{n! * \left ( \dfrac{1}{2} - n \right )}[/math]
That does not make any sense for [imath]a_0[/imath] because j = 1 starts j off above n. So I suspect what is meant is
[math]a_n = \dfrac{\displaystyle \left (\prod_{j=0}^n \dfrac{1}{2} - j \right )}{n! * \left ( \dfrac{1}{2} - n \right )}.[/math]
Let's compute some values.
[math]a_0 = \dfrac{\left ( \dfrac{1}{2} - 0 \right )}{0! * \left ( \dfrac{1}{2} - 0 \right )} = 1.[/math]
[math]a_1 = \dfrac{\left ( \dfrac{1}{2} - 0 \right )\left ( \dfrac{1}{2} - 1 \right )}{1! * \left ( \dfrac{1}{2} - 1 \right )} = \dfrac{\dfrac{1}{2}}{1} = \dfrac{1}{2}[/math]
[math]a_2 = \dfrac{\left ( \dfrac{1}{2} - 0 \right )\left ( \dfrac{1}{2} - 1 \right )\left ( \dfrac{1}{2} - 2 \right )}{2! * \left ( \dfrac{1}{2} - 2 \right )} =\dfrac{\left ( \dfrac{1}{2}\right )\left ( - \dfrac{1}{2} \right)}{2} = - \dfrac{1}{8}[/math]
[math]a_3 = \dfrac{\left ( \dfrac{1}{2} - 0 \right )\left ( \dfrac{1}{2} - 1 \right )\left ( \dfrac{1}{2} - 2 \right )\left ( \dfrac{1}{2} - 3 \right ) }{3! * \left ( \dfrac{1}{2} - 3 \right )} =\\ \dfrac{\left ( \dfrac{1}{2}\right )\left ( - \dfrac{1}{2} \right)\left ( - \dfrac{3}{2} \right)}{3 * 2} = \dfrac{1}{16} [/math]
This looks like a pattern to me. If n is an integer [imath]\ge[/imath] 0, it appears that [imath]a_{n+1} = \text {minus } \dfrac{(2n - 1)a_n}{2n + 2}.[/imath] Can we prove it?
[math]\text {We calculated } a_0 = 1 \text { and } a_1 = \dfrac{1}{2} \text { above}.\\ \therefore a_1 = \dfrac{1}{2} = \dfrac{1}{2} * 1 = \dfrac{1 }{2} * a_0 = \dfrac{(1 - 2 * 0)a_0}{2 * 0 + 2} \implies\\ a_1 = \text {minus } \dfrac{(2 * 0 - 1)a_0}{2 * 0 + 2}.[/math]
Pattern if n = 1. Now consider any integer k [imath]\ge[/imath] 0 such that
[math]a_{k+1} = \text {minus } * \dfrac{(2k - 1)a_k}{2k + 2}.\\ \text {By definition, } a_{(k+1)+1} = \dfrac{\displaystyle \left ( \prod_{j=0}^{(k+1)+1} \dfrac{1}{2} - j \right ) }{\{(k+1) + 1\}! * \left ( \dfrac{1}{2} - \{(k + 1) + 1\} \right )} =\\ \dfrac{\displaystyle \left ( \prod_{j=0}^{k+1} \dfrac{1}{2} - j \right ) * \cancel{\left ( \dfrac{1}{2} - \{(k + 1) + 1\} \right )} }{\{(k+1) + 1\}! * \cancel{\left ( \dfrac{1}{2} - \{(k + 1) + 1\} \right )}} =\\ \dfrac{\displaystyle \left (\prod_{j=0}^{k+1} \dfrac{1}{2} - j \right ) * \left \{ \dfrac{1}{2} - (k + 1) \right \}}{\{(k+1) + 1\}(k + 1)! * \left \{ \dfrac{1}{2} - (k+1) \right \}} =\\ \dfrac{\dfrac{1}{2} - (k + 1)}{(k+1) + 1} * \dfrac{\displaystyle \left (\prod_{j=0}^{k+1} \dfrac{1}{2} - j \right )}{(k + 1)! * \left \{ \dfrac{1}{2} - (k+1) \right \}}.\\ \text {But, by definition, } \dfrac{\displaystyle \left (\prod_{j=0}^{k+1} \dfrac{1}{2} - j \right )}{(k + 1)! * \left \{ \dfrac{1}{2} - (k+1) \right \}} = a_{k+1}.\\ \therefore a_{(k+1)+1} = \dfrac{\dfrac{1}{2} - (k + 1)}{(k+1) + 1} * a_{k+1} = \dfrac{\{1 - 2(k + 1)\}a_{k+1}}{2(k+1) + 2} \implies\\ a_{(k+1)+1} = \text {minus } \dfrac{\{2(k + 1) - 1\}a_{k+1}}{2(k+1) + 2}.[/math]
So we can eliminate the product elements and just focus on the sum. Moreover, we have telescoping possibilities in the sum.
[math]a_n = \dfrac{\dfrac{1}{2} * \displaystyle \prod_{j=1}^n \left ( \dfrac{1}{2} - j \right )}{n! * \left ( \dfrac{1}{2} - n \right )}[/math]
That does not make any sense for [imath]a_0[/imath] because j = 1 starts j off above n. So I suspect what is meant is
[math]a_n = \dfrac{\displaystyle \left (\prod_{j=0}^n \dfrac{1}{2} - j \right )}{n! * \left ( \dfrac{1}{2} - n \right )}.[/math]
Let's compute some values.
[math]a_0 = \dfrac{\left ( \dfrac{1}{2} - 0 \right )}{0! * \left ( \dfrac{1}{2} - 0 \right )} = 1.[/math]
[math]a_1 = \dfrac{\left ( \dfrac{1}{2} - 0 \right )\left ( \dfrac{1}{2} - 1 \right )}{1! * \left ( \dfrac{1}{2} - 1 \right )} = \dfrac{\dfrac{1}{2}}{1} = \dfrac{1}{2}[/math]
[math]a_2 = \dfrac{\left ( \dfrac{1}{2} - 0 \right )\left ( \dfrac{1}{2} - 1 \right )\left ( \dfrac{1}{2} - 2 \right )}{2! * \left ( \dfrac{1}{2} - 2 \right )} =\dfrac{\left ( \dfrac{1}{2}\right )\left ( - \dfrac{1}{2} \right)}{2} = - \dfrac{1}{8}[/math]
[math]a_3 = \dfrac{\left ( \dfrac{1}{2} - 0 \right )\left ( \dfrac{1}{2} - 1 \right )\left ( \dfrac{1}{2} - 2 \right )\left ( \dfrac{1}{2} - 3 \right ) }{3! * \left ( \dfrac{1}{2} - 3 \right )} =\\ \dfrac{\left ( \dfrac{1}{2}\right )\left ( - \dfrac{1}{2} \right)\left ( - \dfrac{3}{2} \right)}{3 * 2} = \dfrac{1}{16} [/math]
This looks like a pattern to me. If n is an integer [imath]\ge[/imath] 0, it appears that [imath]a_{n+1} = \text {minus } \dfrac{(2n - 1)a_n}{2n + 2}.[/imath] Can we prove it?
[math]\text {We calculated } a_0 = 1 \text { and } a_1 = \dfrac{1}{2} \text { above}.\\ \therefore a_1 = \dfrac{1}{2} = \dfrac{1}{2} * 1 = \dfrac{1 }{2} * a_0 = \dfrac{(1 - 2 * 0)a_0}{2 * 0 + 2} \implies\\ a_1 = \text {minus } \dfrac{(2 * 0 - 1)a_0}{2 * 0 + 2}.[/math]
Pattern if n = 1. Now consider any integer k [imath]\ge[/imath] 0 such that
[math]a_{k+1} = \text {minus } * \dfrac{(2k - 1)a_k}{2k + 2}.\\ \text {By definition, } a_{(k+1)+1} = \dfrac{\displaystyle \left ( \prod_{j=0}^{(k+1)+1} \dfrac{1}{2} - j \right ) }{\{(k+1) + 1\}! * \left ( \dfrac{1}{2} - \{(k + 1) + 1\} \right )} =\\ \dfrac{\displaystyle \left ( \prod_{j=0}^{k+1} \dfrac{1}{2} - j \right ) * \cancel{\left ( \dfrac{1}{2} - \{(k + 1) + 1\} \right )} }{\{(k+1) + 1\}! * \cancel{\left ( \dfrac{1}{2} - \{(k + 1) + 1\} \right )}} =\\ \dfrac{\displaystyle \left (\prod_{j=0}^{k+1} \dfrac{1}{2} - j \right ) * \left \{ \dfrac{1}{2} - (k + 1) \right \}}{\{(k+1) + 1\}(k + 1)! * \left \{ \dfrac{1}{2} - (k+1) \right \}} =\\ \dfrac{\dfrac{1}{2} - (k + 1)}{(k+1) + 1} * \dfrac{\displaystyle \left (\prod_{j=0}^{k+1} \dfrac{1}{2} - j \right )}{(k + 1)! * \left \{ \dfrac{1}{2} - (k+1) \right \}}.\\ \text {But, by definition, } \dfrac{\displaystyle \left (\prod_{j=0}^{k+1} \dfrac{1}{2} - j \right )}{(k + 1)! * \left \{ \dfrac{1}{2} - (k+1) \right \}} = a_{k+1}.\\ \therefore a_{(k+1)+1} = \dfrac{\dfrac{1}{2} - (k + 1)}{(k+1) + 1} * a_{k+1} = \dfrac{\{1 - 2(k + 1)\}a_{k+1}}{2(k+1) + 2} \implies\\ a_{(k+1)+1} = \text {minus } \dfrac{\{2(k + 1) - 1\}a_{k+1}}{2(k+1) + 2}.[/math]
So we can eliminate the product elements and just focus on the sum. Moreover, we have telescoping possibilities in the sum.