What is the sum from n=3 to infinity of ne^{-n} ?

The sum from n=3 to infinity of ne^{-n} is converges

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sum from n=3 to infinity of ne^{-n}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{5}{n ^{2} + 49}

n = 0 \sum \infty \frac{n}{n ^{2} - 2 n + 1}

n = 0 \sum \infty \frac{7}{n} - \frac{7}{n + 1}

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

n = 0 \sum \infty \frac{1}{2 + n}

What is the sum from n=3 to infinity of ne^{-n} ?
The sum from n=3 to infinity of ne^{-n} is converges