What is the sum from n=0 to infinity of 1/(9+n) ?

The sum from n=0 to infinity of 1/(9+n) is diverges

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sum from n=0 to infinity of 1/(9+n)

n = 0 \sum \infty \frac{1}{9 + n}

diverges

Solution steps

n = 0 \sum \infty \frac{1}{9 + n}

Apply Series Integral Test:

diverges

= diverges

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

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

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

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

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

What is the sum from n=0 to infinity of 1/(9+n) ?
The sum from n=0 to infinity of 1/(9+n) is diverges