What is the sum from n=1 to infinity of n/(e^{7n)} ?

The sum from n=1 to infinity of n/(e^{7n)} is converges

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sum from n=1 to infinity of n/(e^{7n)}

n = 1 \sum \infty \frac{n}{e ^{7 n}}

converges

Solution steps

n = 1 \sum \infty \frac{n}{e ^{7 n}}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{3 ^{n}}{6 ^{n}}

n = 1 \sum \infty 8 (0.1)^{n - 1}

n = 0 \sum \infty 4 + \frac{5}{n ^{3}}

n = 1 \sum \infty \frac{1}{4 n ^{2} - 49}

n = 1 \sum \infty 6 n^{2} e^{- n^{3}}

What is the sum from n=1 to infinity of n/(e^{7n)} ?
The sum from n=1 to infinity of n/(e^{7n)} is converges