What is the sum from n=2 to infinity of ne^{-5n} ?

The sum from n=2 to infinity of ne^{-5n} is converges

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sum from n=2 to infinity of ne^{-5n}

n = 2 \sum \infty n e^{- 5 n}

converges

Solution steps

n = 2 \sum \infty n e^{- 5 n}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty e^{3 n}

n = 0 \sum \infty \frac{3 ^{n}}{n ^{2}}

n = 1 \sum \infty \frac{3 n}{n ^{2} + 3}

n = 1 \sum \infty (\frac{5}{n})^{n}

k = 1 \sum \infty \frac{4}{k} + 8^{- k}

What is the sum from n=2 to infinity of ne^{-5n} ?
The sum from n=2 to infinity of ne^{-5n} is converges