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

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

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

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

converges

Solution steps

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

Apply Alternating Series Test:

converges

= converges

n = 1 \sum \infty \frac{9}{n ( n + 9 )}

n = 1 \sum \infty e^{an}

n = 1 \sum \infty n^{4} e^{- n}

n = 1 \sum \infty \frac{1}{n ^{\frac{7}{6}}}

n = 0 \sum \infty 10 + 2 n

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