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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= - converges

= converges

n = 1 \sum \infty (\frac{6}{9})^{n - 1}

n = 0 \sum \infty 3 (1 - n)^{- 2}

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

n = 0 \sum \infty (\frac{4}{11})^{n}

j = 1 \sum \infty 2 (- \frac{3}{5})^{j}

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