What is the sum from n=9 to infinity of e^{3-2n} ?

The sum from n=9 to infinity of e^{3-2n} is converges

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sum from n=9 to infinity of e^{3-2n}

n = 9 \sum \infty e^{3 - 2 n}

converges

Solution steps

n = 9 \sum \infty e^{3 - 2 n}

Simplify

e^{3 - 2 n} : e^{3} e^{- 2 n}

= n = 9 \sum \infty e^{3} e^{- 2 n}

Apply the constant multiplication rule:

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

= e^{3} \cdot n = 9 \sum \infty e^{- 2 n}

Apply Series Ratio Test:

converges

= e^{3} converges

= converges

\frac{1}{1 + x}

\int 5 x - 3 d x

x \to 0 lim (e^{π (x + 1)})

y^{^{'}} = 2 x y^{2} + 3 x^{2} y^{2}

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

What is the sum from n=9 to infinity of e^{3-2n} ?
The sum from n=9 to infinity of e^{3-2n} is converges