What is the sum from n=0 to infinity of 1/(4+e^{-n)} ?

The sum from n=0 to infinity of 1/(4+e^{-n)} is diverges

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sum from n=0 to infinity of 1/(4+e^{-n)}

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

y = x^{3} - 3 x + 1, (4, 53)

y = \frac{x}{x + 1}

y^{^{'''}} + 5 y^{^{''}} = 0

\int_{0}^{1} (x - x^{2}) d x

\frac{d}{d x} (- \frac{1}{2 ^{x}})

What is the sum from n=0 to infinity of 1/(4+e^{-n)} ?
The sum from n=0 to infinity of 1/(4+e^{-n)} is diverges