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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty e^{- 7 n}

n = 1 \sum \infty \frac{6}{n ^{2} - 1}

n = 0 \sum \infty \frac{1}{4 - n}

n = 1 \sum \infty \frac{9}{n ( n + 2 )}

n = 1 \sum \infty \frac{4}{( - 3 ) ^{n}}

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