数学题在线解答

问题:判断级数Σ(1!)^2+(2!)^2+(3!)^2+…+(n!)^2/(2n)!的敛散性

结论/结果
收敛
解答

\begin{aligned}
& \sum_{n=1}^{\infin}\frac{(1!)^2+(2!)^2+(3!)^2+…+(n!)^2}{(2n)!}\\
\le& \sum_{n=1}^{\infin}\frac{(n!)^2+(n!)^2+(n!)^2+…+(n!)^2}{(2n)!}\\
=& \sum_{n=1}^{\infin}\frac{n(n!)^2}{(2n)!}\\
=&\sum_{n=1}^{\infin}n \cdot \frac{n}{2n} \cdot \frac{n-1}{2n-1} \cdot \frac{n-2}{2n-2}…\frac{1}{n+1}\\
\le&\sum_{n=1}^{\infin} \frac{1}{2^{n-1}}\\
\le&2
\end{aligned}

从而级数收敛

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