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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

f (x) = sin (2 x^{2})

\int_{0}^{2} 2 π x (1 + \frac{x ^{2}}{2}) d x

x \to 1 lim (x^{2} + 8 x - 2)

\frac{d}{d x} (2 x^{2} cos (2 x))

y = 2 cos (x), at x = \frac{π}{3}

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