What is the sum from n=1 to infinity of 1/(3+4^{-n)} ?

The sum from n=1 to infinity of 1/(3+4^{-n)} is diverges

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sum from n=1 to infinity of 1/(3+4^{-n)}

n = 1 \sum \infty \frac{1}{3 + 4 ^{- n}}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{3 + 4 ^{- n}}

Apply Series Divergence Test:

diverges

= diverges

k = 1 \sum \infty 5 (- \frac{1}{6})^{5 k}

n = 1 \sum \infty \frac{3}{n ^{3} + 7}

n = 1 \sum \infty (\frac{4}{7})^{n - 1}

n = 1 \sum \infty cos (\frac{8}{n})

n = 1 \sum \infty (cos (15))^{n}

What is the sum from n=1 to infinity of 1/(3+4^{-n)} ?
The sum from n=1 to infinity of 1/(3+4^{-n)} is diverges