2
$\begingroup$

take the series $(a_{n})$ defined as $a_n = \log(n)$, what would be a correct argument to show the limit does not exist?

Assume $\lim_{n \rightarrow \infty} \log(n) = L$ Therefore:
$\forall \epsilon >0\space \exists N \space natural \space s.t. \forall n > N , \space |\log(n) - L| < \epsilon$
I have the impression that a proper choice of epsilon dependant on L would allow for some contradiction , any hints on the reasoning are appreciated.

  • 1
    An alternative way (not using the limit definition directly) is to show that $(a_n) $ is not bounded. Then, you can use the result that convergent sequences are bounded (use the contrapositive). – 2017-02-07
  • 0
    @Dave Do you mind elaborating on how showing it is unbounded would be developed? – 2017-02-07
  • 0
    Assume there is some bound $M$ that exists, that is, $\forall n, \log(n) < M$ take some $k > e^M$ we know a natural $n$ exists by the archimedean property of the real numbers. So $\log(k) > M$ which is a contradiction, Therefore the series is unbounded. Is this correct? – 2017-02-07
  • 1
    I suppose. Although you can just make the argument directly (not using contradiction). So just show that you can pick an $n $ such that $\log (n)>M $ for any $M>0$. – 2017-02-07
  • 0
    Let $m$ be any integer greater than $e^n$ with $n\in \mathbb N$. Then $\log m>\log e^n=n$. – 2017-02-07

1 Answers 1

3

Let $\log(x)$ be defined as

$$\log(x)=\int_1^x\frac{1}{t}\,dt$$

for $x>0$. Then, we have for $n>1$

$$\begin{align} \int_1^{2^n} \frac{1}{t} dt&=\int_{1}^{2} \frac{1}{t} \,dt+\int_{2}^{4} \frac{1}{t} dt+\cdots +\int_{2^{n-1}}^{2^{n}} \frac{1}{t} \,dt\\\\ &\ge \int_{1}^{2} \frac{1}{2} \,dt+\int_{2}^{4} \frac{1}{4} \,dt+\cdots +\int_{2^{n-1}}^{2^{n}} \frac{1}{2^n} \,dt\\\\ &=\frac12 (2-1)+\frac14 (4-2) + \cdots + \frac{1}{2^{n}}(2^n-2^{n-1})\\\\ &=\frac{n}{2} \end{align}$$

which tends to infinity as $n \to \infty$.

  • 0
    This is a very interesting approach, thank you! – 2017-02-07
  • 2
    You're welcome! My pleasure. Note that it this proof does not even require that we know that $\log(2^n)=n\log(2)$. However, we can easily show this from the definition using a simple substitution $t=u^n$. [SEE THIS ANSWER](http://math.stackexchange.com/questions/1194447/deriving-properties-of-the-logarithm-from-its-integral-representation/1194478#1194478) to see how we can show that the logarithm function has some of the "well-known" properties directly from the integral definition. – 2017-02-07
  • 0
    This one is really simple. I never thought it could be simpler than $\log 2^{n}=n\log 2$. +1 – 2017-02-07
  • 0
    @ParamanandSingh Thank you! Of course, one needs to prove that $\log(x^n)=n\log(x)$ first. But as I mentioned in the previous comment, this is trivial after making the substitution $t=u^n$. – 2017-02-07