What is the sum from n=1 to infinity of 3ne^{-n} ?

The sum from n=1 to infinity of 3ne^{-n} is converges

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sum from n=1 to infinity of 3ne^{-n}

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 3 converges

= converges

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

n = 1 \sum \infty \frac{1}{n ln ^{3} ( n )}

n = 1 \sum \infty \frac{1}{n ^{2.3}}

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

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

What is the sum from n=1 to infinity of 3ne^{-n} ?
The sum from n=1 to infinity of 3ne^{-n} is converges