What is the sum from n=1 to infinity of 5/n-5/(n^2) ?

The sum from n=1 to infinity of 5/n-5/(n^2) is diverges

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sum from n=1 to infinity of 5/n-5/(n^2)

n = 1 \sum \infty \frac{5}{n} - \frac{5}{n ^{2}}

diverges

Solution steps

n = 1 \sum \infty \frac{5}{n} - \frac{5}{n ^{2}}

Apply Series Integral Test:

diverges

= diverges

n = 0 \sum \infty n - 1

n = 1 \sum \infty (1.075)^{n}

n = 0 \sum \infty (\frac{1}{16})^{n}

n = 1 \sum \infty \frac{1}{n ^{1.03}}

n = 0 \sum \infty sin (3 n)

What is the sum from n=1 to infinity of 5/n-5/(n^2) ?
The sum from n=1 to infinity of 5/n-5/(n^2) is diverges