\[\sum_{n=1}^{\infty }\frac{1}{2n^{2}+4n+\frac{3}{2}}=\sum_{n=1}^{\inft y }\frac{2}{4n^{2}+8n+3}=\\\sum_{n=1}^{\infty }\left ( \frac{1}{2n+1}-\frac{1}{2n+3} \right )=\left ( \frac{1}{3}-\frac{1}{5} \right )+\left ( \frac{1}{5} -\frac{1}{7}\right )+\left ( \frac{1}{7} -\frac{1}{9}\right )+......=\frac{1}{3}\]
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