What is the sum from n=21 to infinity of e^{-0.06n} ?

The sum from n=21 to infinity of e^{-0.06n} is converges

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sum from n=21 to infinity of e^{-0.06n}

n = 21 \sum \infty e^{- 0.06 n}

converges

Solution steps

n = 21 \sum \infty e^{- 0.06 n}

n \to \infty lim (n (\frac{e ^{- 0.06 n}}{e ^{- 0.06 (n + 1)}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

(x^{2} - 9) \frac{d y}{d x} + x y = 0

\int 16 - 4 x^{2} d x

\frac{1}{2} sec (2 x)

\frac{d y}{d x} - \frac{y}{x} = 2 x^{2} ln (x)

x \to 1 lim (4 x^{2} - 2 x + 5)

What is the sum from n=21 to infinity of e^{-0.06n} ?
The sum from n=21 to infinity of e^{-0.06n} is converges