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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

k = 1 \sum \infty \frac{1}{k} - \frac{1}{k + 2}

n = 0 \sum \infty e^{- 19 n}

n = 7 \sum \infty \frac{n !}{n - 6}

k = 1 \sum \infty \frac{1}{k ^{2} + 3 k + 2}

n = 2 \sum \infty \frac{10}{ln ( n )}

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