1

Here's a monstrous sequence I need to find a limit of (or prove it doesn't exist) as $n\rightarrow\infty$.

$$\frac{966\sqrt{n}-1025 n^2+1320n^2\sqrt{n}}{1331\sqrt{n^5}-1410\sqrt[3]{n^4}+1569\sqrt[7]{n^6}}$$

I don't even know where to start. None of the tricks I know don't apply here, changing roots to power of fractions doesn't seem to help, I can't find any division that would simplify finding the limit, etc. Moreover, using integrals or derivatives is disallowed. So where should I start? Any helpful theorems that would help me with solving this?

Jimmy R.
  • 35,868
qiubit
  • 2,313
  • Yuck. Why do you even need to solve this? Are you doing this for amusement/exercise? Did this come up as part of a larger investigation? – Newb Oct 30 '14 at 18:34
  • 2
    Divide top and bottom by their largest power of $n$ – Claude Leibovici Oct 30 '14 at 18:35
  • It's one of "problems" I can have on my nearest exam, and I just had limits on my lecture so I'm not fluent at solving such problems yet :( – qiubit Oct 30 '14 at 18:37
  • Rewrite $\sqrt[7]{n^6}$ as $n^7/6$, and similarly for the others. That should make life easier. Then divide the num. and denom. by the highest power of $n$, and take the limit. – Akiva Weinberger Oct 30 '14 at 18:41

2 Answers2

5

Simply look at the terms of high degree:

$$\frac{966\sqrt{n}-1025 n^2+1320n^2\sqrt{n}}{1331\sqrt{n^5}-1410\sqrt[3]{n^4}+1569\sqrt[7]{n^6}}\sim_\infty\frac{1320n^2\sqrt{n}}{1331\sqrt{n^5}}\xrightarrow{n\to\infty}\frac{1320}{1331}$$

  • Isn't $-1410\sqrt[3]{n^4}$ of higher degree in denominator? – qiubit Oct 30 '14 at 18:39
  • We have $\frac52>\frac43$. –  Oct 30 '14 at 18:40
  • You can reduce that last fraction. Remember, $11^3=1331$. – Ron Gordon Oct 30 '14 at 18:48
  • 1
    You should factor the numerator and the denominator using the highest degree i.e. you write the numerator on this form $$n^2\sqrt n(\frac{966}{n^2}-\frac{1025}{\sqrt n}+1320)$$ and the same for the denominator and then cancel and pass to the limit @user36346 –  Oct 30 '14 at 18:50
  • Yes, sorry, I deleted my comment because I've seen my mistake before you've written @SamiBenRomdhane – qiubit Oct 30 '14 at 18:51
2

The highest power of $n$ in the numerator and denominator is $n^{5/2}$. Dividing top and bottom by that amount, the expression is equal to

$$\frac{966/n^2-1025/n^{1/2}+1320}{1331-1410/n^{7/6}+1569/n^{23/14}}$$

Now it's hopefully clear that in the limit as $n \rightarrow \infty$ what happens, as each term $c/n^{\alpha} \rightarrow 0$ as $\alpha > 0$.

Simon S
  • 26,524