What is the sum from n=1 to infinity of 5/(e^{2n)} ?

The sum from n=1 to infinity of 5/(e^{2n)} is converges

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sum from n=1 to infinity of 5/(e^{2n)}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= 5 converges

= converges

\frac{x y}{x + y}

\int + x^{2} \cdot + sin (x) d x

x^{^{''}} - 3 x^{^{'}} = e^{3 t}

2 cos (x), 4 - 2 cos (x)

\int sin (yz) d y

What is the sum from n=1 to infinity of 5/(e^{2n)} ?
The sum from n=1 to infinity of 5/(e^{2n)} is converges