What is the sum from n=0 to infinity of e^{-3n} ?

The sum from n=0 to infinity of e^{-3n} is converges

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sum from n=0 to infinity of e^{-3n}

n = 0 \sum \infty e^{- 3 n}

converges

Solution steps

n = 0 \sum \infty e^{- 3 n}

Apply Series Integral Test:

converges

= converges

\frac{\partial}{\partial x} (\frac{5 x + 4 y}{4 x - y})

\int_{(1 - x)^{2}}^{(1 + x)^{2}} \int_{1}^{3} y e^{x^{2}} d x d y

x \to - 4 - lim (\frac{3 - 8 x}{x + 4})

\frac{d y}{d x}, 2 (x^{2} + y^{2}) 2 = 81 (x^{2} - y^{2})

\frac{d}{d x} (\frac{x ^{2}}{2} + x \cdot (ln (x) - 1))

What is the sum from n=0 to infinity of e^{-3n} ?
The sum from n=0 to infinity of e^{-3n} is converges