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

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

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

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

converges

Solution steps

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

Simplify

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

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

Apply the constant multiplication rule:

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

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

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

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

Apply Series Ratio Test:

converges

= e converges

= converges

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