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

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

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

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

\frac{e}{e - 1}

Solution steps

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

Apply Series Geometric Test:

\frac{e}{e - 1}

= \frac{e}{e - 1}

(1 - 2) (12)

\int (x^{3} - 5 x)^{5} (3 x^{2} - 5) d x

\frac{d}{d x} ((5 x)^{2})

\frac{20}{s}

y^{^{'}} = 4 y^{2} sin (x)

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