What is the sum from n=1 to infinity of 3/n-3/(n^2) ?

The sum from n=1 to infinity of 3/n-3/(n^2) is diverges

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sum from n=1 to infinity of 3/n-3/(n^2)

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

n = 2 \sum \infty (\frac{3}{4})^{3 n}

n = 1 \sum \infty \frac{5 n}{6 n + 1}

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

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

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

What is the sum from n=1 to infinity of 3/n-3/(n^2) ?
The sum from n=1 to infinity of 3/n-3/(n^2) is diverges