What is the sum from n=1 to infinity of 6/((n+1)!) ?

The sum from n=1 to infinity of 6/((n+1)!) is converges

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sum from n=1 to infinity of 6/((n+1)!)

n = 1 \sum \infty \frac{6}{( n + 1 ) !}

converges

Solution steps

n = 1 \sum \infty \frac{6}{( n + 1 ) !}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{( n + 1 ) !}

Apply Series Ratio Test:

converges

= 6 converges

= converges

y^{^{''}} = a (y^{^{'}})^{2} + b

4 e^{3 x} cos (4 x)

x y

\int \frac{6 x ^{2} - 4 x + 4}{x ^{3} - x ^{2} + 2 x} d x

\frac{d y}{d x} = \frac{y x - y ^{2}}{x ^{2} + y x}

What is the sum from n=1 to infinity of 6/((n+1)!) ?
The sum from n=1 to infinity of 6/((n+1)!) is converges