求下列极限

永远

对被积函数采用二项式定理展开级数求解


\(\displaystyle\mathop {\lim }\limits_{n \to \infty } \int_0^1 {\sqrt[{2n}]{{1 - {x^{2n}}}}} dx\)

永远

考虑到一致收敛，先不谈什么一致收敛，然后，如下所示：


\[\begin{gathered}
  \mathop {\lim }\limits_{n \to \infty } \int_0^1 {\sqrt[{2n}]{{1 - {x^{2n}}}}} dx = \mathop {\lim }\limits_{n \to \infty } \int_0^1 {\sum\limits_m^\infty  {({}_m^{1/2n})( - {x^{2n}})} } dx\quad \quad n \in {N^ * }\quad  \\
  \quad \quad  = \mathop {\lim }\limits_{n \to \infty } \int_0^1 {\sum\limits_{m = 0}^\infty  {({}_m^{1/2n}){{( - 1)}^m}{x^{2nm}}} } dx \\
   = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{m = 0}^\infty  {({}_m^{1/2n}){{( - 1)}^m}\int_0^1 {{x^{2nm}}} } dx \\
   = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{m = 0}^\infty  {({}_m^{1/2n}){{( - 1)}^m}} \frac{1}{{2nm + 1}}\left| {\begin{array}{*{20}{c}}
  1 \\
  0
\end{array}} \right. \\
   = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{m = 0}^\infty  {({}_m^{1/2n}){{( - 1)}^m}} \frac{1}{{2nm + 1}} \\
   = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{m = 0}^\infty  {\frac{{\Gamma (\frac{1}{{2n}} + 1)}}{{\Gamma (m + 1)\Gamma (\frac{1}{{2n}} - m + 1)}}} {( - 1)^m}\frac{1}{{2nm + 1}} \\
\end{gathered} \]

永远

前一段时间学习了陆老师推广的组合数及其gama，beta函数，然后楼主把理论运用到实践中去，结果卡壳了……不知道下面怎么算，郁闷中……


\[\mathop {\lim }\limits_{n \to \infty } \sum\limits_{m = 0}^\infty  {\frac{{\Gamma (\frac{1}{{2n}} + 1)}}{{\Gamma (m + 1)\Gamma (\frac{1}{{2n}} - m + 1)}}} {( - 1)^m}\frac{1}{{2nm + 1}}\]

永远

请教3楼怎么求？