What is the sum from k=1 to infinity of k/(e^{4k)} ?

The sum from k=1 to infinity of k/(e^{4k)} is converges

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sum from k=1 to infinity of k/(e^{4k)}

k = 1 \sum \infty \frac{k}{e ^{4 k}}

converges

Solution steps

k = 1 \sum \infty \frac{k}{e ^{4 k}}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty (\frac{4}{3})^{n}

n = 5 \sum \infty \frac{1}{n ^{2} - 5 n + 6}

n = 1 \sum \infty \frac{1}{n ^{\frac{5}{6}}}

n = 1 \sum \infty \frac{4 ^{n}}{3 ^{n}}

k = 1 \sum \infty (- \frac{3}{5})^{k}

What is the sum from k=1 to infinity of k/(e^{4k)} ?
The sum from k=1 to infinity of k/(e^{4k)} is converges