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

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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

n = 1 \sum \infty (\frac{π}{n})^{n} \cdot n!

n = 10 \sum \infty n

n = 0 \sum \infty n^{2} e^{- 3 n}

n = 2 \sum \infty \frac{6}{n ( n + 2 )}

n = 2 \sum \infty (- 0.45)^{n}

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