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Use the Limit Comparison Test to determine whether the series converges or diverges.\ $sum_{n=1}^{infty} frac{1}{6^n - 3^n}$\ Identify $b_n$ in the following limit.\ $lim_{n o infty} frac{a_n}{b_n} = lim_{n o infty} frac{frac{1}{6^n - 3^n}}{frac{1}{6^n}} = L$\ Since $L$ is a finite number, $L > 0$, and $b_n$ is convergent, the series is convergent

          Use the Limit Comparison Test to determine whether the series converges or diverges.\
$sum_{n=1}^{infty} frac{1}{6^n - 3^n}$\
Identify $b_n$ in the following limit.\
$lim_{n 	o infty} frac{a_n}{b_n} = lim_{n 	o infty} frac{frac{1}{6^n - 3^n}}{frac{1}{6^n}} = L$\
Since $L$ is a finite number, $L > 0$, and $b_n$ is convergent, the series is convergent

$Use the Limit Comparison Test to determine whether the series converges or diverges.sumn=1^infty frac16^n - 3^nIdentify bn in the following limit.limn o infty fracanbn = limn o infty fracfrac16^n - 3^nfrac16^n = LSince L is a finite number, L > 0, and bn is convergent, the series is convergent$

Added by Matthew L.

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