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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{1}{6 n ^{4} + 24 n}

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

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

n = 1 \sum \infty \frac{1}{7 n + 3}

n = 1 \sum \infty \frac{4 ^{n} + n}{n !}

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