What is the sum from n=0 to infinity of (e/(2pi))^n ?

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

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sum from n=0 to infinity of (e/(2pi))^n

n = 0 \sum \infty (\frac{e}{2 π})^{n}

\frac{2 π}{2 π - e}

Solution steps

n = 0 \sum \infty (\frac{e}{2 π})^{n}

Apply Series Geometric Test:

\frac{2 π}{2 π - e}

= \frac{2 π}{2 π - e}

d y = (x y^{2} + 3 x y) d x

y^{^{''}} + 4 y^{^{'}} + 4 y = 25 cos (x)

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

x^{y}

y = x^{2} 2 x - 7

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