What is the sum from n=3 to infinity of 1/(8e^{2n)} ?

The sum from n=3 to infinity of 1/(8e^{2n)} is converges

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sum from n=3 to infinity of 1/(8e^{2n)}

n = 3 \sum \infty \frac{1}{8 e ^{2 n}}

converges

Solution steps

n = 3 \sum \infty \frac{1}{8 e ^{2 n}}

Apply the constant multiplication rule:

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

= \frac{1}{8} \cdot n = 3 \sum \infty \frac{1}{e ^{2 n}}

Apply Series Ratio Test:

converges

= \frac{1}{8} converges

= converges

n = 1 \sum \infty (\frac{1}{7})^{n - 1}

n = 0 \sum \infty \frac{( 0.5 ) ^{n}}{n}

n = 1 \sum \infty 4 (0.9)^{n - 1}

n = 10 \sum \infty \frac{3}{n ^{4.6}}

n = 1 \sum \infty \frac{1}{2 n + 5}

What is the sum from n=3 to infinity of 1/(8e^{2n)} ?
The sum from n=3 to infinity of 1/(8e^{2n)} is converges