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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

\frac{d}{d x} (\frac{1}{e ^{3 x}})

\int \frac{u}{u ^{2} + 4} d u

y = 6 x

y = cos (tan (x))

\int_{0}^{π} 3 sin^{3} (x) d x

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