What is the sum from n=0 to infinity of 1/(4+n) ?

The sum from n=0 to infinity of 1/(4+n) is diverges

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sum from n=0 to infinity of 1/(4+n)

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

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

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

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})

What is the sum from n=0 to infinity of 1/(4+n) ?
The sum from n=0 to infinity of 1/(4+n) is diverges