What is the sum from n=6 to infinity of e^{5-2n} ?

The sum from n=6 to infinity of e^{5-2n} is converges

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sum from n=6 to infinity of e^{5-2n}

n = 6 \sum \infty e^{5 - 2 n}

converges

Solution steps

n = 6 \sum \infty e^{5 - 2 n}

Simplify

e^{5 - 2 n} : e^{5} e^{- 2 n}

= n = 6 \sum \infty e^{5} e^{- 2 n}

Apply the constant multiplication rule:

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

= e^{5} \cdot n = 6 \sum \infty e^{- 2 n}

Apply Series Ratio Test:

converges

= e^{5} converges

= converges

\int 2 + \frac{1}{1 - x} d x

\frac{\partial}{\partial z} (z)

\frac{3}{( x - 1 )}

\int \frac{e ^{x} - 1}{e ^{2 x}} d x

y = - 2 x - 1

What is the sum from n=6 to infinity of e^{5-2n} ?
The sum from n=6 to infinity of e^{5-2n} is converges