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

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

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

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

converges

Solution steps

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

n \to \infty lim (n (\frac{\frac{3 ^{n}}{3 + n !}}{\frac{3 ^{n + 1}}{3 + ( n + 1 ) !}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

n = 1 \sum \infty (\frac{4}{5})^{n - 1}

n = 0 \sum \infty arctan (3 n)

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

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

k = 1 \sum \infty k e^{- 4 k^{2}}

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