2
$\begingroup$

This is my question. I've been told that $1/x^2$ converges while $1/x$ diverges. My intuition tells me that looking at these just plain out as functions that both should converge...my reasoning is as follows...I compare the $x$ and $x^2$ terms and what they mean for the fraction value. Both are decreasing the value of $y$..only thing is the squared term is decreasing it faster. Does this not mean the the plain $x$ term would still decrease the function value to that same value at infinity (I know its not a point) but with more time. All this is taken with respect to improper integrals and computing the area. but that's not relevant to my question...Im just curious about the $y$ values of the respective functions, because I think they should both eventually approach the same number, just the one function approaches it quicker.

$\endgroup$
6
  • 1
    $\begingroup$ Here is meta.math.stackexchange.com/questions/5020/… for properly displaying functions. $\endgroup$
    – Arbuja
    Jan 16, 2016 at 19:32
  • $\begingroup$ The function values do both converge to $0$, but in the case of $\frac1x$, it happens so slowly that the improper integral from $1$ to $\infty$ is infinite, while in the case with $\frac1{x^2}$, the integral is finite. Is that what your asking about? $\endgroup$
    – Arthur
    Jan 16, 2016 at 19:37
  • $\begingroup$ The question doesn't make sense. What do you mean by "$1/x^2$ converges"? $\endgroup$
    – user258700
    Jan 16, 2016 at 19:37
  • $\begingroup$ @AhmedHussein The y value would converge to zero as x approaches infinity $\endgroup$ Jan 16, 2016 at 19:42
  • $\begingroup$ @Arthur Yes that is why im asking. But then with respect to computing an improper integral...even though it happens so slowly...why doesn't the value still occur? Is it because of the mathematical methods used to achieve them...and because you're using ln(x) to find the area under the graph of 1/x? $\endgroup$ Jan 16, 2016 at 19:45

2 Answers 2

2
$\begingroup$

The question is, I believe, why $\int_1^\infty \frac{1}{x}dx$ diverges while $\int_1^\infty \frac{1}{x^2}dx$ converges.

Of course, if we calculate the integrals for both:

$\int_1^\infty \frac{1}{x}dx=lim_{a\rightarrow \infty} ln(x)|_1^a\rightarrow \infty$

$\int_1^\infty \frac{1}{x^2}dx=\lim_{a\rightarrow \infty}-\frac{1}{x}|_1^a=1$

However, this is not the explanation you are looking for! This problem can be related to the sum of the infinite series of $\frac{1}{x}$ and $\frac{1}{x^2}$.

$\sum_{n=1}^k \frac{1}{n}$ is something known as the harmonic series. The truth is, the series eventually diverges as $k\rightarrow \infty$ but it diverges very slowly (the partial sums are of logarithmic growth, as you can see from the integral above).

Wikipedia has a nice example of the counter-intuitive nature of the problem.

$\endgroup$
3
  • $\begingroup$ so even though the denominator of 1/n grows with respect to the sum you showed...it still DIVERGES...I really thought that because the denominator gets larger that it will converge...however if you have the link to why 1/n converges I would really like to see it. Thanks for the help $\endgroup$ Jan 16, 2016 at 20:01
  • $\begingroup$ @TeyashArdun There is a big difference between on one hand, the terms of a sequence or the values of a function going to $0$, and on the other hand, the sum of those terms or the integral of that function being finite. Is it this difference that you have not quite been able to wrap your head around? $\endgroup$
    – Arthur
    Jan 16, 2016 at 20:34
  • $\begingroup$ Yes this is what I need to understand $\endgroup$ Jan 16, 2016 at 21:23
0
$\begingroup$

As you pointed out correctly, one way to define the natural logarithm is as the initegral

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

But how can we prove directly from the definition in $(1)$ that $\lim_{x\to \infty}\log(x)=\infty$?

We do so by examining the behavior of $\log(2^n)$ as $n\to \infty$. Using $(1)$, we write

$$\begin{align} \log(2^n)&=\int_1^{2^n}\frac{1}{t}\,dt\\\\ &=\int_1^2 \frac{1}{t}\,dt +\int_2^4 \frac{1}{t}\,dt+\int_4^8 \frac{1}{t}\,dt+\cdots+\int_{2^{n-1}}^{2^n}\frac{1}{t}\,dt\\\\ &\ge \frac12(2-1)+\frac14(4-2)+\frac18(8-4)+\cdots+\frac{1}{2^n}\left(2^n-2^{n-1}\right)\\\\ &=\frac n2 \end{align}$$

Obviously, as $n\to \infty$, the logarithm tends to $\infty$ and therefore

$$\lim_{x\to \infty}\int_1^x \frac1t \,dt=\infty$$

So, we see that although $\frac1x\to 0$, the area under it grows without bound.

$\endgroup$
2
  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. - Mark $\endgroup$
    – Mark Viola
    Jan 17, 2016 at 18:26
  • $\begingroup$ Shall I delete my answer? $\endgroup$
    – Mark Viola
    Jan 25, 2016 at 3:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .