What is the sum from n=1 to infinity of 1/(n^{2.5)} ?

The sum from n=1 to infinity of 1/(n^{2.5)} is converges

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sum from n=1 to infinity of 1/(n^{2.5)}

n = 1 \sum \infty \frac{1}{n ^{2.5}}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{2.5}}

Apply p-Series Test:

converges

= converges

n = 4 \sum \infty \frac{1}{n ^{2} - 1}

n = 1 \sum \infty \frac{6}{4 n ^{2} - 1}

n = 0 \sum \infty 4 - \frac{1}{n ( n + 1 )}

k = 1 \sum \infty k e^{- 3 k}

n = 1 \sum \infty 2 sin (\frac{2}{n})

What is the sum from n=1 to infinity of 1/(n^{2.5)} ?
The sum from n=1 to infinity of 1/(n^{2.5)} is converges