What is the sum from n=1 to infinity of 1/(sqrt(n)n) ?

The sum from n=1 to infinity of 1/(sqrt(n)n) is converges

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sum from n=1 to infinity of 1/(sqrt(n)n)

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

converges

Solution steps

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

Simplify

n n : n^{\frac{3}{2}}

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

Apply p-Series Test:

converges

= converges

n = 0 \sum \infty - n^{2}

n = 0 \sum \infty 9 e^{- 9 n}

n = 0 \sum \infty 10 (\frac{3}{4})^{n}

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

n = 0 \sum \infty \frac{5}{1 0 ^{n}}

What is the sum from n=1 to infinity of 1/(sqrt(n)n) ?
The sum from n=1 to infinity of 1/(sqrt(n)n) is converges