What is the sum from n=1 to infinity of 3e^{-4n} ?

The sum from n=1 to infinity of 3e^{-4n} is converges

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sum from n=1 to infinity of 3e^{-4n}

n = 1 \sum \infty 3 e^{- 4 n}

converges

Solution steps

n = 1 \sum \infty 3 e^{- 4 n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty e^{- 4 n}

Apply Series Ratio Test:

converges

= 3 converges

= converges

f (x) = \frac{1}{x ^{2} + 4}, (1, \frac{1}{5})

\int 3 e^{x} - 3 x e^{x^{2}} d x

\frac{d}{d x} (k x + d)

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

What is the sum from n=1 to infinity of 3e^{-4n} ?
The sum from n=1 to infinity of 3e^{-4n} is converges