What is the sum from n=1 to infinity of (n!)/(3n!+1) ?

The sum from n=1 to infinity of (n!)/(3n!+1) is diverges

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sum from n=1 to infinity of (n!)/(3n!+1)

n = 1 \sum \infty \frac{n !}{3 n ! + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{n !}{3 n ! + 1}

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty 4 (- \frac{10}{11})^{n}

k = 1 \sum \infty \frac{k}{e ^{4 k}}

n = 0 \sum \infty (\frac{4}{3})^{n}

n = 5 \sum \infty \frac{1}{n ^{2} - 5 n + 6}

n = 1 \sum \infty \frac{1}{n ^{\frac{5}{6}}}

What is the sum from n=1 to infinity of (n!)/(3n!+1) ?
The sum from n=1 to infinity of (n!)/(3n!+1) is diverges