What is the sum from n=1 to infinity of 5/(n+1) ?

The sum from n=1 to infinity of 5/(n+1) is diverges

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sum from n=1 to infinity of 5/(n+1)

n = 1 \sum \infty \frac{5}{n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{5}{n + 1}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{n + 1}

Apply Series Limit Comparison Test:

diverges

= 5 diverges

= diverges

\int_{0}^{1} (3 t - 1)^{50} d t

x \to \infty lim (\frac{x}{4})

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

\frac{\partial}{\partial y} (\frac{7 y}{x ^{5}})

\frac{d}{d x} ((\frac{1 + 3 x}{3 x}) (3 - x))

What is the sum from n=1 to infinity of 5/(n+1) ?
The sum from n=1 to infinity of 5/(n+1) is diverges