What is the sum from n=1 to infinity of 3/(2^n-1) ?

The sum from n=1 to infinity of 3/(2^n-1) is converges

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sum from n=1 to infinity of 3/(2^n-1)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 3 converges

= converges

n = 5 \sum \infty e^{5 - 6 n}

n = 0 \sum \infty \frac{1}{n ^{10}}

n = 1 \sum \infty 2 (- \frac{1}{5})^{2 n}

n = 1 \sum \infty π

n = 1 \sum \infty ln (n - 1)

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