1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ $$n\geqslant2\implies\frac1{n^2}-\frac1n\leqslant-\frac1{2n}$$ $\endgroup$ – Did Oct 6 '15 at 10:10
3
$\begingroup$

Hint: $\sum \frac{1}{n^2}$ converges absolutely and $\sum \frac{1}{n}$ diverges, so what about the sum?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @SyedMuhammadAsad yeah that is correct $\endgroup$ – Dominic Michaelis Oct 18 '15 at 15:51
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

$$\sum\limits_{n=1}^{\infty}{(\frac{1}{n^2}-\frac{1}{n})}:=\lim\limits_{M\to\infty}{\sum\limits_{n=1}^{M}{(\frac{1}{n^2}-\frac{1}{n})}}=\lim\limits_{M\to\infty}{\sum\limits_{n=1}^{M}{\frac{1}{n^2}}}-\lim\limits_{M\to\infty}{\sum\limits_{n=1}^{M}{\frac{1}{n}}}=:\sum\limits_{n=1}^{\infty}{\frac{1}{n^2}}-\sum\limits_{n=1}^{\infty}{\frac{1}{n}}$$ And you know that $\sum\limits_{n=1}^{\infty}{\frac{1}{n}}$ diverges. So the initial series also diverges.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.