What is the sum from n=3 to infinity of e^{5-4n} ?

The sum from n=3 to infinity of e^{5-4n} is converges

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sum from n=3 to infinity of e^{5-4n}

n = 3 \sum \infty e^{5 - 4 n}

converges

Solution steps

n = 3 \sum \infty e^{5 - 4 n}

Apply Series Ratio Test:

converges

= converges

\int \frac{- x}{( x ^{2} + 1 ) ^{2}} d x

x \to - 3 lim (\frac{3 - ∣ x ∣}{x + 3})

y^{^{''}} + 17 y^{^{'}} + 70 y = 0

(- 5, - 2), (3, - 9)

\int_{0}^{2} (10 x - 5 x^{2}) d x

What is the sum from n=3 to infinity of e^{5-4n} ?
The sum from n=3 to infinity of e^{5-4n} is converges