What is the sum from n=1 to infinity of 1/(n^{1.43)} ?

The sum from n=1 to infinity of 1/(n^{1.43)} is converges

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sum from n=1 to infinity of 1/(n^{1.43)}

n = 1 \sum \infty \frac{1}{n ^{1.43}}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{1.43}}

Apply p-Series Test:

converges

= converges

n = 1 \sum \infty \frac{9}{7 n ^{2} + 7 n}

n = 0 \sum \infty \frac{1}{n 2 ^{n}}

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

k = 3 \sum \infty \frac{1}{2 k ^{2} + k}

n = 1 \sum \infty \frac{9}{n ^{\frac{3}{7}}}

What is the sum from n=1 to infinity of 1/(n^{1.43)} ?
The sum from n=1 to infinity of 1/(n^{1.43)} is converges