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

sjeikdom0 2021-09-03 Answered

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

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Expert Answer

falhiblesw
Answered 2021-09-04 Author has 18601 answers

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

converges when

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

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