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I am stuck with the problem on showing that $\{n(n-1)\}$ diverges. I know it leads to the indeterminate form $\infty-\infty$. That is, $$\displaystyle\lim_{n\to\infty}n(n-1)=\displaystyle\lim_{n\to\infty}(n^2-n)=\infty-\infty$$. How do I proceed? Thanks.

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    $\begingroup$ It's also of the form $\infty\cdot \infty$, which is not an ideterminate form. $\endgroup$ Feb 13, 2018 at 1:24
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    $\begingroup$ Be cautious about writing "$=\infty-\infty$." Since $\infty-\infty$ is indeterminate, the limit isn't equal to it. Often you'll lose points on exams for writing this type of statement. $\endgroup$ Feb 13, 2018 at 1:27
  • $\begingroup$ Alternatively, you could write $n=n-1+1$ and get $n(n-1)=(n-1)^2+(n-1)$ which has the form $\infty+\infty$, which is not indeterminate. $\endgroup$ Feb 13, 2018 at 1:33

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For $n> 2$, $n-1 > 1$

Hence we have $n(n-1) > n$

That is $$\lim_{n \to \infty} n(n-1) \ge \lim_{n \to \infty} n = \infty$$

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HINT

Let $a_n = n(n-1)$ for $n \ge 1$. Note that $a_{n+1} > a_n$ and in addition, if $M \in \mathbb{R}$ then $a_{M+1} = M(M+1) > M$, so the sequence is unbounded above.

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