What is the sum from n=1 to infinity of 1/(5+e^{-n)} ?

The sum from n=1 to infinity of 1/(5+e^{-n)} is diverges

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sum from n=1 to infinity of 1/(5+e^{-n)}

n = 1 \sum \infty \frac{1}{5 + e ^{- n}}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{5 + e ^{- n}}

Apply Series Divergence Test:

diverges

= diverges

m = 1 \sum \infty \frac{10}{m ^{8} + 3}

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

n = 1 \sum \infty \frac{n}{n ^{2} + 6}

k = 1 \sum \infty \frac{5}{k} - \frac{5}{k + 1}

n = 1 \sum \infty \frac{5}{n ( n + 1 )}

What is the sum from n=1 to infinity of 1/(5+e^{-n)} ?
The sum from n=1 to infinity of 1/(5+e^{-n)} is diverges