What is the sum from n=1 to infinity of 8/((n+1)!) ?

The sum from n=1 to infinity of 8/((n+1)!) is converges

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sum from n=1 to infinity of 8/((n+1)!)

n = 1 \sum \infty \frac{8}{( n + 1 ) !}

converges

Solution steps

n = 1 \sum \infty \frac{8}{( n + 1 ) !}

Apply the constant multiplication rule:

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

= 8 \cdot n = 1 \sum \infty \frac{1}{( n + 1 ) !}

Apply Series Ratio Test:

converges

= 8 converges

= converges

What is the sum from n=1 to infinity of 8/((n+1)!) ?
The sum from n=1 to infinity of 8/((n+1)!) is converges