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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

x \to \frac{π}{2} lim (\frac{1 - tan ( 2 x )}{2 tan ( x )})

x^{- 8 c o s (x)}

\frac{d}{d x} (4 x^{2} cos (2 x))

\frac{d y}{d x} = x + y - 1

\frac{- 2 e ^{- 2 s}}{( s ^{2} - 2 s + 1 )}

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