What is the sum from n=6 to infinity of e^{3-2n} ?

The sum from n=6 to infinity of e^{3-2n} is converges

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sum from n=6 to infinity of e^{3-2n}

n = 6 \sum \infty e^{3 - 2 n}

converges

Solution steps

n = 6 \sum \infty e^{3 - 2 n}

Simplify

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

= n = 6 \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 = 6 \sum \infty e^{- 2 n}

Apply Series Ratio Test:

converges

= e^{3} converges

= converges

x^{2} y^{^{''}} + x y^{^{'}} = 0

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\int sin^{7} (\frac{x}{9}) cos (\frac{x}{9}) d x

\int ln (3 x + 1) d x

What is the sum from n=6 to infinity of e^{3-2n} ?
The sum from n=6 to infinity of e^{3-2n} is converges