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I have to show that: \begin{eqnarray} \sum_{n=1}^{\infty}(-1)^n\left(\frac{1\cdot 3\cdot 5 \cdot 7...(2n-1)}{2\cdot4\cdot6\cdot8...(2n)}\right)^{3} \end{eqnarray} is absolute convergent.


So I have to prove that: \begin{eqnarray} \sum_{n=1}^{\infty}\left(\frac{1\cdot3\cdot5\cdot7...(2n-1)}{2\cdot4\cdot6\cdot8...(2n)}\right)^{3} \end{eqnarray} is convergent. But I see that I have to use the Gauss's test, so if $a_{n}$ is \begin{eqnarray} a_{n}=\left(\frac{1\cdot3\cdot5\cdot7...(2n-1)}{2\cdot4\cdot6\cdot8...(2n)}\right)^{3} \end{eqnarray}

I have to write: \begin{eqnarray} \frac{a_{n+1}}{a_{n}}= 1 - \frac{A}{n}+\frac{f(n)}{n^{s}} \end{eqnarray}
where $A>1$, $s>1$ and $M>0$ such that $|f(n)|\leq M$, then I compute $\frac{a_{n+1}}{a_{n}}$ like: \begin{eqnarray*} \frac{a_{n+1}}{a_{n}}&=&\left(\frac{2n+1}{2n+2}\right)^{3}\\ &=&\frac{8n^{3}+12n^{2}+6n+1}{8n^{3}+24n^{2}+24n+8}\\ &=&1 + \frac{-12n^{2}-18n-7}{8n^{3}+24n^{2}+24n+8} \end{eqnarray*} But I can't compute the form that I have to get. Do you how can I continue? or How can I express this terms of Gauss's test?

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    $\begingroup$ Perform the long division $$\frac{a_{n+1}}{a_{n}}=\left(\frac{2n+1}{2n+2}\right)^{3}=1-\frac{3}{2 n}+\frac{9}{4 n^2}+O\left(\frac{1}{n^3}\right)$$ $\endgroup$ Oct 2, 2021 at 4:12

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Try the Stirling's approximation:

$$ n! \sim \sqrt{2\pi n} \big(\frac{n}{e}\big)^n. $$

then

\begin{array}{l} \frac{1*3*5*7...(2n-1)}{2*4*6*8...(2n)} \\ = {\left[\frac{2 n !}{(2^n n !)^{2}}\right]} \\ \sim\frac{\sqrt{4 \pi n}\left( \frac{2 n}{e}\right)^{2 n}}{2^{2 n} 2 \pi n\left(\frac{n}{e}\right)^{2n}} \\ \sim \frac{1}{\sqrt{4\pi n}} \end{array}

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