What is the sum from n=1 to infinity of 2ne^{-n} ?

The sum from n=1 to infinity of 2ne^{-n} is converges

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sum from n=1 to infinity of 2ne^{-n}

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

converges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Ratio Test:

converges

= 2 converges

= converges

n = 0 \sum \infty \frac{1}{3 n ^{2}}

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

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

n = 1 \sum \infty \frac{1}{n ^{7} + 2}

k = 0 \sum \infty \frac{k}{1 + k ^{4}}

What is the sum from n=1 to infinity of 2ne^{-n} ?
The sum from n=1 to infinity of 2ne^{-n} is converges