What is the sum from n=8 to infinity of e^{7-6n} ?

The sum from n=8 to infinity of e^{7-6n} is converges

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sum from n=8 to infinity of e^{7-6n}

n = 8 \sum \infty e^{7 - 6 n}

converges

Solution steps

n = 8 \sum \infty e^{7 - 6 n}

Apply Series Ratio Test:

converges

= converges

x y^{^{'}} = 2 x^{2} y + 3 x^{3} e^{x^{2}}, y (0) = 5

- x + 2, x^{2} - 1, [0, 1]

t sin (3 t)

\frac{\partial}{\partial x} (x^{2} + y^{2} - 25)

\frac{d}{d x} (2 x y + z^{2})

What is the sum from n=8 to infinity of e^{7-6n} ?
The sum from n=8 to infinity of e^{7-6n} is converges