What is the sum from n=1 to infinity of 9/n-9/(n^2) ?

The sum from n=1 to infinity of 9/n-9/(n^2) is diverges

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sum from n=1 to infinity of 9/n-9/(n^2)

n = 1 \sum \infty \frac{9}{n} - \frac{9}{n ^{2}}

diverges

Solution steps

n = 1 \sum \infty \frac{9}{n} - \frac{9}{n ^{2}}

Apply Series Integral Test:

diverges

= diverges

n = 3 \sum \infty arctan (n)

n = 1 \sum \infty \frac{1}{( 4 n ) !}

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

n = 0 \sum \infty \frac{5}{0. 5 ^{n} - 2}

n = 0 \sum \infty \frac{2 ^{n}}{n ^{7}}

What is the sum from n=1 to infinity of 9/n-9/(n^2) ?
The sum from n=1 to infinity of 9/n-9/(n^2) is diverges