What is the sum from n=1 to infinity of 1/(9+e^{-n)} ?

The sum from n=1 to infinity of 1/(9+e^{-n)} is diverges

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sum from n=1 to infinity of 1/(9+e^{-n)}

n = 1 \sum \infty \frac{1}{9 + e ^{- n}}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{9 + e ^{- n}}

Apply Series Divergence Test:

diverges

= diverges

n = 2 \sum \infty (- 0.5)^{n}

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

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}

What is the sum from n=1 to infinity of 1/(9+e^{-n)} ?
The sum from n=1 to infinity of 1/(9+e^{-n)} is diverges