What is the sum from n=0 to infinity of e^{(2n-1)} ?

The sum from n=0 to infinity of e^{(2n-1)} is diverges

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sum from n=0 to infinity of e^{(2n-1)}

n = 0 \sum \infty e^{(2 n - 1)}

diverges

Solution steps

n = 0 \sum \infty e^{2 n - 1}

Simplify

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

= n = 0 \sum \infty e^{2 n} e^{- 1}

Apply the constant multiplication rule:

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

= e^{- 1} \cdot n = 0 \sum \infty e^{2 n}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{e} \cdot n = 0 \sum \infty e^{2 n}

Apply Series Ratio Test:

diverges

= \frac{1}{e} diverges

= diverges

\int \frac{2 x + 1}{x ^{2} - x + 3} d x

x \to 2 lim (\frac{8}{x})

\int_{9}^{\infty} \frac{2}{x ^{3}} d x

(y e^{x y} - \frac{1}{y}) d x (x e^{x y} - \frac{x}{y ^{2}}) d y = 0

f (x) = \frac{5 t + 300}{t ^{2} + 25}

What is the sum from n=0 to infinity of e^{(2n-1)} ?
The sum from n=0 to infinity of e^{(2n-1)} is diverges