What is the sum from n=0 to infinity of 1/(3+n) ?

The sum from n=0 to infinity of 1/(3+n) is diverges

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sum from n=0 to infinity of 1/(3+n)

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

n = 1 \sum \infty e^{n^{- 2}}

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

n = 1 \sum \infty \frac{2}{n}

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

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

What is the sum from n=0 to infinity of 1/(3+n) ?
The sum from n=0 to infinity of 1/(3+n) is diverges