Convergence of a series
- Tolaso J Kos
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Convergence of a series
Examine if the series
$$\sum_{n=1}^{\infty} \left [ \frac{1\cdot 3 \cdot 5\cdots \left ( 2n-1 \right )}{2 \cdot 4 \cdot 6 \cdots \left ( 2n \right )} \right ]^2 $$
converges.
$$\sum_{n=1}^{\infty} \left [ \frac{1\cdot 3 \cdot 5\cdots \left ( 2n-1 \right )}{2 \cdot 4 \cdot 6 \cdots \left ( 2n \right )} \right ]^2 $$
converges.
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- Grigorios Kostakos
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- Location: Ioannina, Greece
Re: Convergence of a series
Let $\alpha_n=\big(\frac{(2n-1)!!}{(2n)!!}\big)^2\,,\; n\in\mathbb{N}$ and $\beta_n=n\,,\; n\in\mathbb{N}$. Then $\sum_{n=1}^{\infty}\frac{1}{\beta_n}=+\infty$ and for all $n\in\mathbb{N}$ holds
\begin{align*}
\beta_n-\beta_{n+1}\,\frac{\alpha_{n+1}}{\alpha_n}&=-\frac{1}{4(n+1)}<0\,.
\end{align*}
By Kummer's criterion we have that $\sum_{n=1}^{\infty}\big(\frac{(2n-1)!!}{(2n)!!}\big)^2=+\infty$.
\begin{align*}
\beta_n-\beta_{n+1}\,\frac{\alpha_{n+1}}{\alpha_n}&=-\frac{1}{4(n+1)}<0\,.
\end{align*}
By Kummer's criterion we have that $\sum_{n=1}^{\infty}\big(\frac{(2n-1)!!}{(2n)!!}\big)^2=+\infty$.
Grigorios Kostakos
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