What is the sum from n=1 to infinity of 100e^{-n/2} ?

The sum from n=1 to infinity of 100e^{-n/2} is converges

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sum from n=1 to infinity of 100e^{-n/2}

n = 1 \sum \infty 100 e^{- \frac{n}{2}}

converges

Solution steps

n = 1 \sum \infty 100 e^{- \frac{n}{2}}

Apply the constant multiplication rule:

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

= 100 \cdot n = 1 \sum \infty e^{- \frac{n}{2}}

Simplify

e^{- \frac{n}{2}} : \frac{1}{e ^{\frac{n}{2}}}

= 100 \cdot n = 1 \sum \infty \frac{1}{e ^{\frac{n}{2}}}

Apply Series Ratio Test:

converges

= 100 converges

= converges

f (x) = (3 - 2 x^{3})^{2}

f (x) = x^{2} + 1

\int - \frac{3}{4} e^{- 4 x} x^{2} d x

y = x - x^{3} (1)

\frac{d y}{d x} + y^{9} x + 8 y = 0

What is the sum from n=1 to infinity of 100e^{-n/2} ?
The sum from n=1 to infinity of 100e^{-n/2} is converges