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I need to simplify the expression of this series: $$ \sum\limits_{n=1}^{\infty}\frac{3^{n}(4n)!}{1 · 4 · 7 · ⋯ ·3n+1} $$ As the ratio (for the test): $$\frac{a_{n+1}}{a_n}$$

First I multiplied the reciprocal of a(n) by a(n+1):

$$ \frac{a_{n+1}}{a_n}=\frac{3^{n+1}(4(n+1))!}{(3(n+1)+1)}\cdot\frac{3n+1}{3^{n}(4n)!} $$

Then I expanded the terms:

$$ \frac{3^{n+1}\cdot4n!(4n+4)(4n+3)(4n+2)(4n+1)(3n+1)}{3^{n}\cdot4n!(3n+4)(3n+3)(3n+2)(3n+1)} $$

Simplified down to:

$$ \frac{a_{n+1}}{a_n}=\frac{3(4n+4)(4n+3)(4n+2)(4n+1)}{(3n+4)(3n+3)(3n+2)} $$

Am I missing a step or performing any incorrectly?

This question is from a practice test, which is multiple choice, but I don't have an answer key: See choices here

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  • $\begingroup$ I think the term $3n+1$ in the numerator should not be there... $\endgroup$
    – Miguel
    Jun 30, 2017 at 9:54
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    $\begingroup$ Why did you wrote $(3(n+1)+1)=(3n+4)(3n+3)(3n+2)(3n+1)$ in your denominator? That is wrong and the right answer is amongst the given. It should be $3(n+1)+1=3n+3+1=3n+4$. $\endgroup$ Jun 30, 2017 at 9:57
  • $\begingroup$ What is the point of simplifying the terms of a clearly divergent series? $3^n(4n)!$ grows incredibly faster than $(3n+1)!!!$. $\endgroup$ Jun 30, 2017 at 15:23

2 Answers 2

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\begin{align} \frac{a_{n+1}}{a_n}&= \frac{3^{n+1}\cdot4n!(4n+4)(4n+3)(4n+2)(4n+1)(3n+1)}{3^{n}\cdot 4n!(3n+4)}\\ &= \frac{3 (4n+4)(4n+3)(4n+2)(4n+1)(3n+1)}{(3n+4)}. \end{align} Answer F.

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As I saw, the thing you are missing is in the product of denominators,

In product there is a difference of $3$ in consecutive terms , so when dividing denominator of $a_{n+1}~by ~a_n$ It should be $3n+4$ which is left,because $3n+4$ is the consecutive of $3n+1$ not $(3n+2)(3n+3)(3n+4)$

Numerator is all good , so the answer should be $$\frac{3 (4n+4)(4n+3)(4n+2)(4n+1)(3n+1)}{(3n+4)}$$ which is option F and how is shown by @PaoloLeonetti

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  • $\begingroup$ I had a feeling I messed up the denominator. I mistakenly unpacked it like a factorial. Thanks. $\endgroup$ Jun 30, 2017 at 15:59

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