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 diverges

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

n = 1 \sum \infty \frac{4}{n + 1}

diverges

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 Limit Comparison Test:

diverges

= 4 diverges

= diverges

n = 0 \sum \infty n (\frac{4}{5})^{n}

\frac{\partial}{\partial x} (cos (2 x + 2 y))

n = 1 \sum \infty - \frac{3}{2 ^{n}}

\int \frac{x ^{2}}{13 - 4 x + x ^{2}} d x

\frac{- 3 x - 16 x}{2 x ^{\frac{1}{3}}}

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 diverges