proof of mean square convergence of the sample mean of a stationary process

n\\operatorname{var}(\\bar{X}_{n})=\\frac{1}{n}\\sum_{i=1}^{n}\\sum_{j=1}^{n}%\\operatorname{cov}(X_{i},X_{j})\\\\=\\sum_{|h|<n}(1-\\frac{|h|}{n})\\gamma(h)\\leq\\sum_{|h|<n}|\\gamma(h)|

If $\\gamma(n)\\to 0$ as $n\\to\\infty$ then $\\lim_{n\\to\\infty}\\frac{1}{n}\\sum_{|h|<n}|\\gamma(h)|=2\\lim_{n\\to\\infty}|\\gamma(%n)|=0$ , whence $\\operatorname{var}[\\bar{X}_{n}]\\to 0$ .
If $\\sum_{h=-\\infty}^{\\infty}|\\gamma(h)|<\\infty$ then the dominated Convergence theorem gives

\\lim_{n\\to\\infty}\\sum_{|h|<n}(1-\\frac{|h|}{n})\\gamma(h)=\\sum_{h=-\\infty}^{%\\infty}\\gamma(h)