Find simplified form of ratio: a_{n+1}/a_n w/ a_n = [5n (4n)!] / [2*4*6*...*(2n + 1)]

[calculusIsKillingMe]

Ratio Test Calculus Question Please Help:

Consider the following series:

. . . . .\(\displaystyle \displaystyle \sum_{n = 1}^{\infty}\, \dfrac{5n\, (4n)!}{2\, \cdot\, 4\, \cdot\, 6\, \cdot\, ...\, \cdot\, (2n\, +\, 2)}\)

Find a simplified expression for the ratio:

. . . . .\(\displaystyle \dfrac{a_{n+1}}{a_n}\)

One area of this question that I find particularly confusing is the 2*4*6... I thought it could be represented as 2n! but that does not seem to be the case

 

[Deleted member 4993]

Ratio Test Calculus Question Please Help:

Consider the following series:

. . . . .\(\displaystyle \displaystyle \sum_{n = 1}^{\infty}\, \dfrac{5n\, (4n)!}{2\, \cdot\, 4\, \cdot\, 6\, \cdot\, ...\, \cdot\, (2n\, +\, 2)}\)

Find a simplified expression for the ratio:

. . . . .\(\displaystyle \dfrac{a_{n+1}}{a_n}\)

One area of this question that I find particularly confusing is the 2*4*6... I thought it could be represented as 2n! but that does not seem to be the case

2*4*6...(2n+2) = 2*1 * 2*2 * 2*3 *... * 2*(n+1) = [2*2*2 .... *(2)] * [1*2*3*... (n+1)] = ???

This is really simple algebraic manipulation!

What are your thoughts?

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[stapel]

Ratio Test Calculus Question Please Help:

Consider the following series:

. . . . .\(\displaystyle \displaystyle \sum_{n = 1}^{\infty}\, \dfrac{5n\, (4n)!}{2\, \cdot\, 4\, \cdot\, 6\, \cdot\, ...\, \cdot\, (2n\, +\, 2)}\)

Find a simplified expression for the ratio:

. . . . .\(\displaystyle \dfrac{a_{n+1}}{a_n}\)

One area of this question that I find particularly confusing is the 2*4*6... I thought it could be represented as 2n! but that does not seem to be the case

When the pattern is given to you, explicitly, as "2*4*6*...*(2n + 2)", why would you think that the pattern would be "2*4*6*...*(2n)"? :shock:

 

[Deleted member 4993]

Originally Posted by calculusIsKillingMe Ratio Test Calculus Question Please Help:

Consider the following series:

. . . . .[FONT=MathJax_Size2]∑[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]∞[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]![/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main])[/FONT]



Find a simplified expression for the ratio:

. . . . .[FONT=MathJax_Math]a[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Math]n[/FONT]

One area of this question that I find particularly confusing is the 2*4*6... I thought it could be represented as 2n! but that does not seem to be the case

2*4*6...(2n+2) = 2*1 * 2*2 * 2*3 *... * 2*(n+1) = [2*2*2 .... *(2)] * [1*2*3*... (n+1)] = ???

No response from OP in 48 hrs.

2*4*6...(2n+2) = 2*1 * 2*2 * 2*3 *... * 2*(n+1) = [2*2*2 .... *(2)] * [1*2*3*... (n+1)] = 2⁽ⁿ⁺¹⁾ * {(n+1)!}

a_n = [FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]![/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main])[/FONT]
\(\displaystyle \displaystyle{a_n \ = \ \dfrac{5*n*(4*n!)}{2^{n+1}(n+1)!}}\)

\(\displaystyle \displaystyle{a_{n+1} \ = \ \dfrac{5 * (n+1) * [4 * (n+1)!]}{2^{n+2}(n+2)!}}\)

\(\displaystyle \displaystyle{\dfrac{a_{n+1}}{a_n} \ = \ \dfrac{5 * (n+1) * [4 * (n+1)!]}{5*n*(4*n!)} * \dfrac{2^{n+1}(n+1)!}{2^{n+2}(n+2)!}}\)

\(\displaystyle \displaystyle{\dfrac{a_{n+1}}{a_n} \ = \ \dfrac{(n+1)^2}{n} * \dfrac{1}{2(n+2)}}\)

 

[Steven G]

Ratio Test Calculus Question Please Help:

Consider the following series:

. . . . .\(\displaystyle \displaystyle \sum_{n = 1}^{\infty}\, \dfrac{5n\, (4n)!}{2\, \cdot\, 4\, \cdot\, 6\, \cdot\, ...\, \cdot\, (2n\, +\, 2)}\)

Find a simplified expression for the ratio:

. . . . .\(\displaystyle \dfrac{a_{n+1}}{a_n}\)

One area of this question that I find particularly confusing is the 2*4*6... I thought it could be represented as 2n! but that does not seem to be the case

If n=4, then (2n)!= 8!= 1*2*3*4*5*6*7*8. As already pointed out, you can simply factor out the 2's. For the record, you do know that basic algebra is a pre-requiste for calculus?