What is the sum from n=1 to infinity of ne^{-8n} ?

The sum from n=1 to infinity of ne^{-8n} is converges

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sum from n=1 to infinity of ne^{-8n}

n = 1 \sum \infty n e^{- 8 n}

converges

Solution steps

n = 1 \sum \infty n e^{- 8 n}

Apply Series Ratio Test:

converges

= converges

tan^{4} (x)

\frac{\partial}{\partial x} ((2 + 3 x) (3 - 2 x))

y^{^{'}} = \frac{( x - y )}{( x + 2 y )}

\int \frac{1}{x ^{2}} ln (x) d x

\frac{d}{d y} (\frac{ln ( y )}{2 y} + \frac{1}{y})

What is the sum from n=1 to infinity of ne^{-8n} ?
The sum from n=1 to infinity of ne^{-8n} is converges