What is the sum from n=3 to infinity of (2/11)^{-n} ?

The sum from n=3 to infinity of (2/11)^{-n} is diverges

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sum from n=3 to infinity of (2/11)^{-n}

n = 3 \sum \infty (\frac{2}{11})^{- n}

diverges

Solution steps

n = 3 \sum \infty (\frac{2}{11})^{- n}

Simplify

(\frac{2}{11})^{- n} : \frac{1}{( \frac{2}{11} ) ^{n}}

= n = 3 \sum \infty \frac{1}{( \frac{2}{11} ) ^{n}}

Simplify

\frac{1}{( \frac{2}{11} ) ^{n}} : (\frac{11}{2})^{n}

= n = 3 \sum \infty (\frac{11}{2})^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty (\frac{11}{2})^{n} - (\frac{11}{2})^{0} - (\frac{11}{2})^{1} - (\frac{11}{2})^{2}

n = 0 \sum \infty (\frac{11}{2})^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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\int \frac{1}{25 - x ^{2}} d x

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f^{^{'}} (x) = 0.7 sin (x) - 1.1 cos (x)

What is the sum from n=3 to infinity of (2/11)^{-n} ?
The sum from n=3 to infinity of (2/11)^{-n} is diverges