What is the sum from n=1 to infinity of n/(e^{n^4)} ?

The sum from n=1 to infinity of n/(e^{n^4)} is converges

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sum from n=1 to infinity of n/(e^{n^4)}

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

converges

Solution steps

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

Apply Series Root Test:

converges

= converges

n = 1 \sum \infty 4 (0.5)^{n - 1}

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

n = 1 \sum \infty \frac{1}{16 n ^{2} - 8 n - 3}

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

n = 1 \sum \infty \frac{5}{( - 4 ) ^{n}}

What is the sum from n=1 to infinity of n/(e^{n^4)} ?
The sum from n=1 to infinity of n/(e^{n^4)} is converges