What is the sum from n=1 to infinity of 7/n-7/(n^2) ?

The sum from n=1 to infinity of 7/n-7/(n^2) is diverges

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sum from n=1 to infinity of 7/n-7/(n^2)

n = 1 \sum \infty \frac{7}{n} - \frac{7}{n ^{2}}

diverges

Solution steps

n = 1 \sum \infty \frac{7}{n} - \frac{7}{n ^{2}}

Apply Series Integral Test:

diverges

= diverges

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

n = 1 \sum \infty (0.8)^{n}

n = 1 \sum \infty (\frac{3}{7})^{n}

k = 1 \sum \infty 2^{- k}

k = 5 \sum \infty (\frac{e}{π})^{k - 1}

What is the sum from n=1 to infinity of 7/n-7/(n^2) ?
The sum from n=1 to infinity of 7/n-7/(n^2) is diverges