What is the sum from n=6 to infinity of e^{3-4n} ?

The sum from n=6 to infinity of e^{3-4n} is converges

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sum from n=6 to infinity of e^{3-4n}

n = 6 \sum \infty e^{3 - 4 n}

converges

Solution steps

n = 6 \sum \infty e^{3 - 4 n}

Apply Series Ratio Test:

converges

= converges

x^{2} x + 1

\frac{d}{d x} ((x^{5} + 5)^{\frac{4}{3}} tan (2 x))

x \to \infty lim (4 x - x^{2})

\int 2 x e^{9 x} d x

\frac{\partial}{\partial y} (x^{\frac{3}{2}} - 3 y)

What is the sum from n=6 to infinity of e^{3-4n} ?
The sum from n=6 to infinity of e^{3-4n} is converges