What is the sum from n=1 to infinity of 1/(3+e^{-n)} ?

The sum from n=1 to infinity of 1/(3+e^{-n)} is diverges

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sum from n=1 to infinity of 1/(3+e^{-n)}

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty \frac{99}{10 0 ^{n}}

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

n = 3 \sum \infty n^{2} e^{- n^{3}}

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

n = 1 \sum \infty \frac{n !}{7 ^{n}}

What is the sum from n=1 to infinity of 1/(3+e^{-n)} ?
The sum from n=1 to infinity of 1/(3+e^{-n)} is diverges