What is the sum from n=2 to infinity of e^{5-6n} ?

The sum from n=2 to infinity of e^{5-6n} is converges

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sum from n=2 to infinity of e^{5-6n}

n = 2 \sum \infty e^{5 - 6 n}

converges

Solution steps

n = 2 \sum \infty e^{5 - 6 n}

Simplify

e^{5 - 6 n} : e^{5} e^{- 6 n}

= n = 2 \sum \infty e^{5} e^{- 6 n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= e^{5} \cdot n = 2 \sum \infty e^{- 6 n}

Apply Series Ratio Test:

converges

= e^{5} converges

= converges

\frac{d ^{3}}{d x ^{3}} (2 x)

\int_{1}^{3} 9 arctan (1 x) d x

f (x) = x^{3} 1 - x^{2}

y = 1 + x, y = \frac{( x + 3 )}{3}

\int sec^{3} (9 x) d x

What is the sum from n=2 to infinity of e^{5-6n} ?
The sum from n=2 to infinity of e^{5-6n} is converges