What is the sum from n=1 to infinity of (n+8)/(n^2) ?

The sum from n=1 to infinity of (n+8)/(n^2) is diverges

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sum from n=1 to infinity of (n+8)/(n^2)

n = 1 \sum \infty \frac{n + 8}{n ^{2}}

diverges

Solution steps

n = 1 \sum \infty \frac{n + 8}{n ^{2}}

Apply Series Comparison Test:

diverges

= diverges

n = 0 \sum \infty \frac{i ^{2 n}}{4 n}

n = 1 \sum \infty n (\frac{1}{7})^{n}

n = 1 \sum \infty \frac{( 3 n - 2 )}{n}

n = 2 \sum \infty (\frac{2}{5})^{n}

n = 1 \sum \infty \frac{4 ^{n}}{n ^{7}}

What is the sum from n=1 to infinity of (n+8)/(n^2) ?
The sum from n=1 to infinity of (n+8)/(n^2) is diverges