What is the sum from n=4 to infinity of e^{5-6n} ?

The sum from n=4 to infinity of e^{5-6n} is converges

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sum from n=4 to infinity of e^{5-6n}

n = 4 \sum \infty e^{5 - 6 n}

converges

Solution steps

n = 4 \sum \infty e^{5 - 6 n}

Apply Series Ratio Test:

converges

= converges

\frac{\partial}{\partial y} (y (1 + ln (x y)))

t y^{^{'}} + t y = 1 - y, y (1) = 0

\int \frac{1}{x - 1} d x

(x^{3} - 5 x) (4 x^{2} - 3 x)

x + k \cdot x

What is the sum from n=4 to infinity of e^{5-6n} ?
The sum from n=4 to infinity of e^{5-6n} is converges