What is the sum from n=5 to infinity of e^{3-2n} ?

The sum from n=5 to infinity of e^{3-2n} is converges

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sum from n=5 to infinity of e^{3-2n}

n = 5 \sum \infty e^{3 - 2 n}

converges

Solution steps

n = 5 \sum \infty e^{3 - 2 n}

Simplify

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

= n = 5 \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 = 5 \sum \infty e^{- 2 n}

Apply Series Ratio Test:

converges

= e^{3} converges

= converges

\int 3 t^{2} sin (2 t^{3} + 1) d t

\int \frac{2 x}{x} d x

n = 1 \sum \infty \frac{17 n !}{n ^{n}}

y^{2} = 4 x, y = 0, y = 4

(3 x^{2} + y + 3 x^{3} y) d x + (x) d y = 0

What is the sum from n=5 to infinity of e^{3-2n} ?
The sum from n=5 to infinity of e^{3-2n} is converges