What is the sum from n=0 to infinity of e^{3-2n} ?

The sum from n=0 to infinity of e^{3-2n} is converges

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sum from n=0 to infinity of e^{3-2n}

n = 0 \sum \infty e^{3 - 2 n}

converges

Solution steps

n = 0 \sum \infty e^{3 - 2 n}

Simplify

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

= n = 0 \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 = 0 \sum \infty e^{- 2 n}

Apply Series Ratio Test:

converges

= e^{3} converges

= converges

x \to 0 lim (\frac{a ^{2 x} - 1}{x})

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

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

\int \frac{x}{x ^{2} + 8 x} d x

y = a x^{3}

What is the sum from n=0 to infinity of e^{3-2n} ?
The sum from n=0 to infinity of e^{3-2n} is converges