What is the sum from n=1 to infinity of 1/(6+e^{-n)} ?

The sum from n=1 to infinity of 1/(6+e^{-n)} is diverges

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sum from n=1 to infinity of 1/(6+e^{-n)}

n = 1 \sum \infty \frac{1}{6 + e ^{- n}}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{6 + e ^{- n}}

Apply Series Divergence Test:

diverges

= diverges

n = 2 \sum \infty \frac{2}{n ln ( n )}

n = 1 \sum \infty 5^{- 3 n}

n = 1 \sum \infty \frac{1}{e ^{4 n}}

n = 2 \sum \infty (- 0.55)^{n}

n = 1 \sum \infty \frac{2}{n ^{3} + 4}

What is the sum from n=1 to infinity of 1/(6+e^{-n)} ?
The sum from n=1 to infinity of 1/(6+e^{-n)} is diverges