What is the sum from n=0 to infinity of 5/(e^{2n)} ?

The sum from n=0 to infinity of 5/(e^{2n)} is converges

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sum from n=0 to infinity of 5/(e^{2n)}

n = 0 \sum \infty \frac{5}{e ^{2 n}}

converges

Solution steps

n = 0 \sum \infty \frac{5}{e ^{2 n}}

Apply the constant multiplication rule:

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

= 5 \cdot n = 0 \sum \infty \frac{1}{e ^{2 n}}

Apply Series Ratio Test:

converges

= 5 converges

= converges

\frac{d}{d x} (e^{x - 1} - 1)

\frac{d y}{d x} = x^{8} y

f (x) = 1 (- 5 x + 3) (- 3 x + 2)

\int \frac{1}{7 + 2 x} d x

f (X) = 2 (\frac{X ^{2}}{20} - 2)^{2} + 2

What is the sum from n=0 to infinity of 5/(e^{2n)} ?
The sum from n=0 to infinity of 5/(e^{2n)} is converges