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

The sum from n=1 to infinity of (pi/e)^{-n} is converges

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

n = 1 \sum \infty (\frac{π}{e})^{- n}

converges

Solution steps

n = 1 \sum \infty (\frac{π}{e})^{- n}

Simplify

(\frac{π}{e})^{- n} : \frac{1}{( \frac{π}{e} ) ^{n}}

= n = 1 \sum \infty \frac{1}{( \frac{π}{e} ) ^{n}}

Simplify

\frac{1}{( \frac{π}{e} ) ^{n}} : (\frac{e}{π})^{n}

= n = 1 \sum \infty (\frac{e}{π})^{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{e}{π})^{n} - (\frac{e}{π})^{0}

n = 0 \sum \infty (\frac{e}{π})^{n} =

converges

= converges - (\frac{e}{π})^{0}

= converges

n = 3 \sum \infty n

n = 1 \sum \infty cos (7 n)

n = 1 \sum \infty \frac{4}{n ^{5}}

n = 0 \sum \infty \frac{3 n}{n ^{2} - 7}

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

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