What is the sum from n=9 to infinity of e^{3-6n} ?

The sum from n=9 to infinity of e^{3-6n} is converges

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sum from n=9 to infinity of e^{3-6n}

n = 9 \sum \infty e^{3 - 6 n}

converges

Solution steps

n = 9 \sum \infty e^{3 - 6 n}

Apply Series Ratio Test:

converges

= converges

(y^{2} - 1) d x - (2 y + x y) d y = 0

y = x^{2} - 3 x + 1

\frac{\partial}{\partial x} (4 (x - y) e^{4 y + 4 x^{2}})

x^{^{''}} + 3 x^{^{'}} - 10 x = 6 e^{4 t}

\frac{\partial}{\partial x} (2 y x^{2} - 4)

What is the sum from n=9 to infinity of e^{3-6n} ?
The sum from n=9 to infinity of e^{3-6n} is converges