What is the sum from n=0 to infinity of 4/(n+3) ?

The sum from n=0 to infinity of 4/(n+3) is diverges

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sum from n=0 to infinity of 4/(n+3)

n = 0 \sum \infty \frac{4}{n + 3}

diverges

Solution steps

n = 0 \sum \infty \frac{4}{n + 3}

Apply the constant multiplication rule:

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

= 4 \cdot n = 0 \sum \infty \frac{1}{n + 3}

Apply Series Integral Test:

diverges

= 4 diverges

= diverges

n = 0 \sum \infty (\frac{n}{n + 1})^{n}

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

n = 0 \sum \infty sin (\frac{nπ}{2})

n = 1 \sum \infty \frac{1}{n ^{2} + 4 n}

n = 0 \sum \infty \frac{1}{100 + n}

What is the sum from n=0 to infinity of 4/(n+3) ?
The sum from n=0 to infinity of 4/(n+3) is diverges