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I've found some other pretty similar questions, but I have to find the limit:

$$\lim_{n\to \infty} \frac{\log n}{n} $$

using the fact that $$\log n = \int_1^n \frac{1}{x} dx $$

and $x^{-1}\leq x^{-\frac{1}{2}}$ if $x>1$.

Finding the limit in some way by l'Hopital or comparison, etc. isn't that difficult, however, we have to use the integral part since we have to use is in other questions.

Can you give me a hint?

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    $\begingroup$ $\displaystyle \log n = \int_1^n \frac{1}{x}\,dx \le \int_1^n \frac{1}{\sqrt{x}}\,dx = 2\sqrt{n} - 2$ amounts to a comparison test. $\endgroup$ – r9m Jan 9 '15 at 17:12
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    $\begingroup$ $ \begin{array}{l} \frac{1}{n} = m \\ \mathop {\lim }\limits_{n \to + \infty } \frac{{\log n}}{n} = \mathop {\lim }\limits_{m \to 0^ + } m\log \frac{1}{m} = - \mathop {\lim }\limits_{m \to 0^ + } m\log m = 0 \\ \end{array} $ $\endgroup$ – sabachir Jan 9 '15 at 19:12
  • $\begingroup$ @r9m I think your comment is a nice answer (probably the clearest, shortest and most elegant). Ponder to post it assuch. $\endgroup$ – Timbuc Jan 10 '15 at 0:06
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$$if,n>1 \\\log(n) \leq \sqrt{n}\\ $$so $$0\leq\lim_{n \to \infty }\frac{logn}{n}<\lim_{n \to \infty } \frac{\sqrt{n}}{n}\rightarrow 0$$

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  • $\begingroup$ I believe this is not what the OP requested. $\endgroup$ – Aaron Maroja Jan 9 '15 at 17:18
  • $\begingroup$ I managed that answer as well, however in the succeeding question, I have to prove that for $\alpha>0, \frac{\log n}{n^\alpha} \to 0 $ when $n \to \infty$, which is the complicated part in fact. $\endgroup$ – sillystudent Jan 9 '15 at 17:20
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Let $ a>0$, then for $n\ge 1$ you have

$$0\le \frac {\ln n}{n^a}=\frac{\int_1^n\frac {dx}{x}}{n^a}\le \frac{\int_1^n\frac {dx}{x^{1-a/2}}}{n^a} = \frac 2a \frac{ {n^{a/2}-1}}{n^a}\to0\quad\text{as}\quad {n\to\infty}$$

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Let $x=\frac{1}{n}$. Then, $$\lim_\limits{n\to +\infty}{x}=0^+$$ and $$\lim_\limits{n\to +\infty}{\frac{\log n}{n}}=\lim_\limits{x\to 0^+}{(x\cdot \log \frac{1}{x})}=-\lim_\limits{x\to 0^+}{(x\cdot \log x)}=0$$

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