What is the sum from n=1 to infinity of 1/(2+e^{-n)} ?

The sum from n=1 to infinity of 1/(2+e^{-n)} is diverges

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sum from n=1 to infinity of 1/(2+e^{-n)}

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

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

n = 0 \sum \infty \frac{6 ^{n}}{n !}

n = 1 \sum \infty n^{2} e^{- 2 n}

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

n = 0 \sum \infty (1 - n)^{- 2}

What is the sum from n=1 to infinity of 1/(2+e^{-n)} ?
The sum from n=1 to infinity of 1/(2+e^{-n)} is diverges