What is the sum from n=3 to infinity of (3/11)^{-n} ?

The sum from n=3 to infinity of (3/11)^{-n} is diverges

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sum from n=3 to infinity of (3/11)^{-n}

n = 3 \sum \infty (\frac{3}{11})^{- n}

diverges

Solution steps

n = 3 \sum \infty (\frac{3}{11})^{- n}

Simplify

(\frac{3}{11})^{- n} : \frac{1}{( \frac{3}{11} ) ^{n}}

= n = 3 \sum \infty \frac{1}{( \frac{3}{11} ) ^{n}}

Simplify

\frac{1}{( \frac{3}{11} ) ^{n}} : (\frac{11}{3})^{n}

= n = 3 \sum \infty (\frac{11}{3})^{n}

Apply Series Ratio Test:

diverges

= diverges

\int \frac{8}{x ln ( 2 x )} d x

\frac{\partial}{\partial x} (e^{- x y})

x \to + 2 + lim (x^{2})

\frac{d}{d x} (z arctan (\frac{y}{x}))

\frac{e ^{x}}{e ^{x} + 1}

What is the sum from n=3 to infinity of (3/11)^{-n} ?
The sum from n=3 to infinity of (3/11)^{-n} is diverges