# Does the series $\sum{(\frac{1}{n^2} - \frac{1}{n})}$ converge or diverge?

Whats about the given series, Converge or Diverge? The series is given $$\sum_1^\infty{(\frac{1}{n^2} - \frac{1}{n})}$$

• This is a good example showing the reader that the comparison test may not apply when the general term of a series is negative. – Megadeth Oct 6 '15 at 9:32
• $$n\geqslant2\implies\frac1{n^2}-\frac1n\leqslant-\frac1{2n}$$ – Did Oct 6 '15 at 10:10

Hint: $\sum \frac{1}{n^2}$ converges absolutely and $\sum \frac{1}{n}$ diverges, so what about the sum?
If the series were convergent assume the sum is $l$. This would mean all the terms being positive that
$$\sum_{n=1}^\infty{1\over n}=-l+{\pi^2\over 6}$$
$$\sum\limits_{n=1}^{\infty}{(\frac{1}{n^2}-\frac{1}{n})}:=\lim\limits_{M\to\infty}{\sum\limits_{n=1}^{M}{(\frac{1}{n^2}-\frac{1}{n})}}=\lim\limits_{M\to\infty}{\sum\limits_{n=1}^{M}{\frac{1}{n^2}}}-\lim\limits_{M\to\infty}{\sum\limits_{n=1}^{M}{\frac{1}{n}}}=:\sum\limits_{n=1}^{\infty}{\frac{1}{n^2}}-\sum\limits_{n=1}^{\infty}{\frac{1}{n}}$$ And you know that $\sum\limits_{n=1}^{\infty}{\frac{1}{n}}$ diverges. So the initial series also diverges.