What is the sum from n=1 to infinity of 1/(n^{1.03)} ?

The sum from n=1 to infinity of 1/(n^{1.03)} is converges

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sum from n=1 to infinity of 1/(n^{1.03)}

n = 1 \sum \infty \frac{1}{n ^{1.03}}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{1.03}}

Apply p-Series Test:

converges

= converges

n = 0 \sum \infty sin (3 n)

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

n = 1 \sum \infty (\frac{4}{9})^{n - 1}

n = 0 \sum \infty (3.5)^{- n}

n = 1 \sum \infty \frac{10 n}{4 n + 1}

What is the sum from n=1 to infinity of 1/(n^{1.03)} ?
The sum from n=1 to infinity of 1/(n^{1.03)} is converges