What is the sum from n=0 to infinity}e^{-1sqrt(n of) ?

The sum from n=0 to infinity}e^{-1sqrt(n of) is converges

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sum from n=0 to infinity}e^{-1sqrt(n of)

n = 0 \sum \infty e^{- 1 n}

converges

Solution steps

n = 0 \sum \infty e^{- 1 \cdot n}

Simplify

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

= n = 0 \sum \infty \frac{1}{e ^{n}}

Apply Series Integral Test:

converges

= converges

n = 1 \sum \infty \frac{2 n}{7 n - 6}

n = 1 \sum \infty n + 3

n = 0 \sum \infty \frac{7}{5 n + 4}

n = 1 \sum \infty 5 (\frac{1}{3})^{n - 1}

n = 0 \sum \infty \frac{3}{n !}

What is the sum from n=0 to infinity}e^{-1sqrt(n of) ?
The sum from n=0 to infinity}e^{-1sqrt(n of) is converges