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

sjeikdom0

sjeikdom0

Answered question

2021-09-03

x=1n+6n2+12n+8

Answer & Explanation

falhiblesw

falhiblesw

Skilled2021-09-04Added 97 answers

x=1n+6n2+12n+8

converges when

n+6n(n+12)+8=0

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