What is the sum from k=1 to infinity of k/(e^{3k)} ?

The sum from k=1 to infinity of k/(e^{3k)} is converges

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sum from k=1 to infinity of k/(e^{3k)}

k = 1 \sum \infty \frac{k}{e ^{3 k}}

converges

Solution steps

k = 1 \sum \infty \frac{k}{e ^{3 k}}

Apply Series Ratio Test:

converges

= converges

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

n = 0 \sum \infty \frac{- 3 ^{n}}{3 ^{n}}

n = 1 \sum \infty 4^{n} 5^{1 - n}

n = 1 \sum \infty (- \frac{1}{7})^{n}

n = 1 \sum \infty sin (\frac{πn}{2})

What is the sum from k=1 to infinity of k/(e^{3k)} ?
The sum from k=1 to infinity of k/(e^{3k)} is converges