What is the sum from n=0 to infinity of (-1+1/e)^n ?

The sum from n=0 to infinity of (-1+1/e)^n is e/(-1+2e)

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sum from n=0 to infinity of (-1+1/e)^n

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

\frac{e}{- 1 + 2 e}

Solution steps

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

Apply Series Geometric Test:

\frac{e}{- 1 + 2 e}

= \frac{e}{- 1 + 2 e}

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What is the sum from n=0 to infinity of (-1+1/e)^n ?
The sum from n=0 to infinity of (-1+1/e)^n is e/(-1+2e)