Whats about the given series, Converge or Diverge? The series is given $$ \sum_1^\infty{(\frac{1}{n^2} - \frac{1}{n})} $$
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$\begingroup$ This is a good example showing the reader that the comparison test may not apply when the general term of a series is negative. $\endgroup$ – Megadeth Oct 6 '15 at 9:32
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1$\begingroup$ $$n\geqslant2\implies\frac1{n^2}-\frac1n\leqslant-\frac1{2n}$$ $\endgroup$ – Did Oct 6 '15 at 10:10
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Hint: $\sum \frac{1}{n^2}$ converges absolutely and $\sum \frac{1}{n}$ diverges, so what about the sum?
answered Oct 6 '15 at 8:36
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$\begingroup$ Please tell me that. 1/n^2 converges because power of n is greater than 1 and 1/n diverge because power of n is equal to 1. $\endgroup$ – Syed Muhammad Asad Oct 18 '15 at 12:58
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1$\begingroup$ @SyedMuhammadAsad yeah that is correct $\endgroup$ – Dominic Michaelis Oct 18 '15 at 15:51
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If the series were convergent assume the sum is $l$. This would mean all the terms being positive that
$$\sum_{n=1}^\infty{1\over n}=-l+{\pi^2\over 6}$$
But the harmonic series is divergent so the initial series is divergent
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$$\sum\limits_{n=1}^{\infty}{(\frac{1}{n^2}-\frac{1}{n})}:=\lim\limits_{M\to\infty}{\sum\limits_{n=1}^{M}{(\frac{1}{n^2}-\frac{1}{n})}}=\lim\limits_{M\to\infty}{\sum\limits_{n=1}^{M}{\frac{1}{n^2}}}-\lim\limits_{M\to\infty}{\sum\limits_{n=1}^{M}{\frac{1}{n}}}=:\sum\limits_{n=1}^{\infty}{\frac{1}{n^2}}-\sum\limits_{n=1}^{\infty}{\frac{1}{n}}$$ And you know that $\sum\limits_{n=1}^{\infty}{\frac{1}{n}}$ diverges. So the initial series also diverges.
