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

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

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

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

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

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

n = 1 \sum \infty \frac{2}{3 n + 8}

n = 1 \sum \infty \frac{1}{11 n ^{3} + 5}

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

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