What is the sum from n=1 to infinity of 2(-e)^{-n} ?

The sum from n=1 to infinity of 2(-e)^{-n} is converges

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sum from n=1 to infinity of 2(-e)^{-n}

n = 1 \sum \infty 2 (- e)^{- n}

converges

Solution steps

n = 1 \sum \infty 2 (- e)^{- n}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty (- e)^{- n}

Apply Series Ratio Test:

converges

= 2 converges

= converges

\int \frac{x ^{5}}{x ^{2} + 7} d x

\int \frac{1}{5 + 4 x - x ^{2}} d x

x \to - \infty lim (\frac{( 4 x ^{2} + 1 )}{( 2 - 7 x )})

x y^{^{'}} + 4 y = - 25 x sin (x^{5})

\frac{d}{d x} (\frac{3}{7 x ^{4} - 5 x ^{2} + 1})

What is the sum from n=1 to infinity of 2(-e)^{-n} ?
The sum from n=1 to infinity of 2(-e)^{-n} is converges