What is the sum from n=5 to infinity of (-1)^ne^{-n} ?

The sum from n=5 to infinity of (-1)^ne^{-n} is converges

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sum from n=5 to infinity of (-1)^ne^{-n}

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

converges

Solution steps

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

Apply Alternating Series Test:

converges

= converges

n = 1 \sum \infty \frac{1 + ln ( n )}{n}

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

n = 1 \sum \infty \frac{n !}{8 4 ^{n}}

n = 2 \sum \infty \frac{4}{ln ( n ^{2} )}

n = 0 \sum \infty \frac{1}{1 + 8 n}

What is the sum from n=5 to infinity of (-1)^ne^{-n} ?
The sum from n=5 to infinity of (-1)^ne^{-n} is converges