Comparing the Indefinite Integrals Convergence for $1/x$ and $1/x^{2}$ between 1 and $\infty$. 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.
 A: 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. 
A: 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.
