What is the sum from n=1 to infinity of 8/(e^n-1) ?

The sum from n=1 to infinity of 8/(e^n-1) is converges

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sum from n=1 to infinity of 8/(e^n-1)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 8 converges

= converges

n = 0 \sum \infty \frac{1}{2 n ^{3}}

n = 0 \sum \infty \frac{n}{( n + 1 ) ^{2}}

n = 0 \sum \infty \frac{n + 1}{3 n + 1}

n = 0 \sum \infty e^{- 14 n}

n = 1 \sum \infty \frac{n}{2 n ^{2} + 11}

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