What is the sum from n=0 to infinity of n/(e^{7n)} ?

The sum from n=0 to infinity of n/(e^{7n)} is converges

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sum from n=0 to infinity of n/(e^{7n)}

n = 0 \sum \infty \frac{n}{e ^{7 n}}

converges

Solution steps

n = 0 \sum \infty \frac{n}{e ^{7 n}}

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{n}{e ^{7 n}} = (n) e^{- 7 n}

= n = 0 \sum \infty n e^{- 7 n}

Apply Series Ratio Test:

converges

= converges

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

n = 2 \sum \infty \frac{5}{n ^{3}}

n = 0 \sum \infty \frac{( 0.8 ) ^{n}}{7}

n = 1 \sum \infty (\frac{n}{2 n})^{n}

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

What is the sum from n=0 to infinity of n/(e^{7n)} ?
The sum from n=0 to infinity of n/(e^{7n)} is converges