What is the sum from n=3 to infinity of e^{5-6n} ?

The sum from n=3 to infinity of e^{5-6n} is converges

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sum from n=3 to infinity of e^{5-6n}

n = 3 \sum \infty e^{5 - 6 n}

converges

Solution steps

n = 3 \sum \infty e^{5 - 6 n}

Apply Series Ratio Test:

converges

= converges

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = 0

tan (x) - 1 sec (x)

\int y^{2} + 1 d y

y = sin (sin (x)), at x = π

What is the sum from n=3 to infinity of e^{5-6n} ?
The sum from n=3 to infinity of e^{5-6n} is converges