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

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

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

n = 1 \sum \infty (\frac{- e}{π})^{n}

\frac{π}{π + e} - 1

Solution steps

n = 1 \sum \infty (\frac{- e}{π})^{n}

Simplify

= 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} = \frac{π}{π + e}

= \frac{π}{π + e} - (- \frac{e}{π})^{0}

Simplify

= \frac{π}{π + e} - 1

n = 0 \sum \infty 3 (- 1)^{n}

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

n = 3 \sum \infty \frac{4}{n ^{2} - 4}

k = 1 \sum \infty \frac{1}{k + 3}

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

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