What is the sum from n=1 to infinity of 1/(8+e^{-n)} ?

The sum from n=1 to infinity of 1/(8+e^{-n)} is diverges

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sum from n=1 to infinity of 1/(8+e^{-n)}

n = 1 \sum \infty \frac{1}{8 + e ^{- n}}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{8 + e ^{- n}}

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty \frac{1}{n ^{2} + 4}

n = 0 \sum \infty \frac{1}{1 + 2 n ^{2}}

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

n = 1 \sum \infty \frac{4}{n !}

n = 1 \sum \infty \frac{n ^{9}}{9 ^{n}}

What is the sum from n=1 to infinity of 1/(8+e^{-n)} ?
The sum from n=1 to infinity of 1/(8+e^{-n)} is diverges