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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Ratio Test:

converges

= 3 converges

= converges

x \to 0 lim (\frac{3}{x} - \frac{3}{x ^{2} + x})

4

y^{^{'}} = \frac{3 x ^{6} e ^{\frac{y}{x}} + x ^{4} y ^{2}}{x ^{5} y}

\frac{d}{d x} ((x + 1)^{\frac{1}{2}})

\int ln^{5} (x) d x

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