What is the sum from n=1 to infinity of 4/((n+1)!) ?

The sum from n=1 to infinity of 4/((n+1)!) is converges

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sum from n=1 to infinity of 4/((n+1)!)

n = 1 \sum \infty \frac{4}{( n + 1 ) !}

converges

Solution steps

n = 1 \sum \infty \frac{4}{( n + 1 ) !}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty \frac{1}{( n + 1 ) !}

Apply Series Ratio Test:

converges

= 4 converges

= converges

f (x) = 7 x^{2}

\frac{5}{4} x^{\frac{4}{3}}

f (x) = ln (x - 2)

y = 2 cos (x + y)

\frac{d ^{2}}{d x ^{2}} (x^{2} - y^{2})

What is the sum from n=1 to infinity of 4/((n+1)!) ?
The sum from n=1 to infinity of 4/((n+1)!) is converges