What is the sum from n=2 to infinity of e^{3-4n} ?

The sum from n=2 to infinity of e^{3-4n} is converges

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sum from n=2 to infinity of e^{3-4n}

n = 2 \sum \infty e^{3 - 4 n}

converges

Solution steps

n = 2 \sum \infty e^{3 - 4 n}

Simplify

e^{3 - 4 n} : e^{3} e^{- 4 n}

= n = 2 \sum \infty e^{3} e^{- 4 n}

Apply the constant multiplication rule:

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

= e^{3} \cdot n = 2 \sum \infty e^{- 4 n}

Apply Series Ratio Test:

converges

= e^{3} converges

= converges

f (x) = (2 x - 8) 4 (x^{2} + x + 1) 5

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\frac{6}{7 t ^{4}}

What is the sum from n=2 to infinity of e^{3-4n} ?
The sum from n=2 to infinity of e^{3-4n} is converges