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

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

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

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

converges

Solution steps

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

Simplify

e^{- n} : \frac{1}{e ^{n}}

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

Simplify

n \frac{1}{e ^{n}} : \frac{n}{e ^{n}}

= n = 2 \sum \infty \frac{n}{e ^{n}}

Apply Series Ratio Test:

converges

= converges

f (x) = e^{4 a} (4 a^{3} - 3)

\frac{d}{d x} (ln (tan (x)))

\int 7 x^{2} x^{3} + 15 d x

\frac{d y}{d x} = 3 x + y

9 x - 9 x

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