计算极限：1/2((n+1)^(2H)+(n-1)^(2H)-2n^(2H))/(H(2H-1)n^(2H-2))

首先，我们可以将分子化简：

\begin{aligned}
&rac{1}{2}\left(\frac{(n+1)^{2H}}{n^{2H-2}}+\frac{(n-1)^{2H}}{n^{2H-2}}-\frac{2n^{2H}}{n^{2H-2}}\right)\
=&\frac{1}{2}\left(\left(\frac{n+1}{n}\right)^{2H}+\left(\frac{n-1}{n}\right)^{2H}-2\right)\
=&\frac{1}{2}\left[\left(1+\frac{1}{n}\right)^{2H}+\left(1-\frac{1}{n}\right)^{2H}-2\right]
\end{aligned}

接下来，我们将分母中的$n^{2H-2}$移到分式的前面：

\begin{aligned}
&\frac{1}{H(2H-1)n^{2H-2}}\
=&\frac{\frac{1}{n^{2H}}}{H(2H-1)n^{-2}}\
=&\frac{1}{H(2H-1)n^{-2}}\cdot\frac{1}{n^{2H}}\
=&\frac{1}{H(2H-1)}\left(\frac{1}{n^2}\right)^{H-1}
\end{aligned}

现在，我们可以将分式的分子和分母带入极限公式：

\begin{aligned}
\lim_{n\to\infty}\frac{\frac{1}{2}\left[\left(1+\frac{1}{n}\right)^{2H}+\left(1-\frac{1}{n}\right)^{2H}-2\right]}{\frac{1}{H(2H-1)}\left(\frac{1}{n^2}\right)^{H-1}} &=\lim_{n\to\infty}\frac{\frac{1}{2}\left[\left(1+\frac{1}{n}\right)^{2H}+\left(1-\frac{1}{n}\right)^{2H}-2\right]}{\frac{1}{H(2H-1)}\cdot\frac{1}{n^{2H-2H+1}}}\
&=\lim_{n\to\infty}\frac{\frac{1}{2}\left[\left(1+\frac{1}{n}\right)^{2H}+\left(1-\frac{1}{n}\right)^{2H}-2\right]\cdot n^{2H-1}}{\frac{1}{H(2H-1)}}\
&=\frac{1}{2}\cdot H(2H-1)\cdot 2\
&=\boxed{H(2H-1)}
\end{aligned}