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}{3 ^{n} n}

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}

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