What is the sum from n=0 to infinity of (n+1)/(n^3) ?

The sum from n=0 to infinity of (n+1)/(n^3) is converges

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sum from n=0 to infinity of (n+1)/(n^3)

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

converges

Solution steps

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

Expand

\frac{n + 1}{n ^{3}} : \frac{1}{n ^{2}} + \frac{1}{n ^{3}}

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

Apply the Sum Rule:

\sum a_{n} + b_{n} = \sum a_{n} + \sum b_{n}

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

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

converges

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

converges

= converges + converges

= converges

n = 1 \sum \infty \frac{7}{6 n}

k = 1 \sum \infty tan (\frac{1}{k})

n = 1 \sum \infty \frac{1}{25 n ^{2} - 1}

k = 1 \sum \infty k (\frac{1}{3})^{k}

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

What is the sum from n=0 to infinity of (n+1)/(n^3) ?
The sum from n=0 to infinity of (n+1)/(n^3) is converges