What is the sum from n=1 to infinity of 1/(7+e^{-n)} ?

The sum from n=1 to infinity of 1/(7+e^{-n)} is diverges

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sum from n=1 to infinity of 1/(7+e^{-n)}

n = 1 \sum \infty \frac{1}{7 + e ^{- n}}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{7 + e ^{- n}}

Apply Series Divergence Test:

diverges

= diverges

n \to \infty lim (3^{n} 7^{- n})

x \to 9 - lim (x^{2} + 5 x + 7)

y^{^{''}} + 12 y^{^{'}} + 36 = 0, y (1) = 0, y^{^{'}} (1) = 1

y^{^{'}} + y = (x + 1)^{2}

\int \frac{10}{x ^{2} + 1} d x

What is the sum from n=1 to infinity of 1/(7+e^{-n)} ?
The sum from n=1 to infinity of 1/(7+e^{-n)} is diverges