# {\sum_{{{x}={1}}}^{\infty}}\frac{{{n}+{6}}}{{{n}^{2}+{12}{n}+{8}}}

$${\sum_{{{x}={1}}}^{\infty}}\frac{{{n}+{6}}}{{{n}^{2}+{12}{n}+{8}}}$$

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falhiblesw

$${\sum_{{{x}={1}}}^{\infty}}\frac{{{n}+{6}}}{{{n}^{2}+{12}{n}+{8}}}$$

converges when

$$\frac{{{n}+{6}}}{{{n}{\left({n}+{12}\right)}+{8}}}={0}$$