Possible proof(?!): $\int_1^\infty \frac{1}{x} dx$ converges I'm probably wrong, as I am not a calculus professor. But I hear that the 
$$\int_1^\infty \frac{1}{x} dx$$
does NOT converge. because $\ln |x|$ approaches infinity but:
We know that the $\lim_{x \to +\infty} \frac{1}{x}$ converges to $0$. As $x$ gets bigger and bigger, $\frac{1}{x}$ gets smaller and smaller.
Also, by definition, the integration notation tells us the area under the curve $f(x)$ and above $y=0$ (which is just the $x$-axis). The area is always between those two lines.
The squeeze theorem tells us that the height of $\frac{1}{x}$ (or the positive distance from $y=0$) > $y$-coordinate of area (the height of the area) > $y=0$ And if the $$\lim_{x \to +\infty} \frac{1}{x} = 0  > \text{ height of area above }y=0$$ then the height of the area under the curve must also equal $0$, and converge.
If the height of the area under the curve $= 0$ then the area at that point must also equal $0$. (you can think of a rectangle with height $0$, it's just a line at that point with height $0$, therefore no area). So if at some point the area equals $0$, then the integral (area under the curve) must converge to a finite number.
 A: It appears that you’re confusing what happens in the limit with what happens at actual real numbers. There is no value of $x$ for which $\frac1x=0$: no matter what positive real number $x$ may be, $\frac1x>0$. Thus, there is no point at which the height of the curve is $0$. The statement that $\lim\limits_{x\to\infty}\frac1x=0$ does not mean that $\frac1x$ is $0$ at some point; it just means that you can get $\frac1x$ to be as small as you want if you take $x$ large enough.
It is in fact true that if you look at chunks of area under the curve that are, say, one unit wide, their areas get closer and closer to $0$ as you move to the right: $\int_n^{n+1}\frac1x\,dx$ gets smaller as $n$ increases, and by taking $n$ large enough, you can make the area, $\int_n^{n+1}\frac1x\,dx$, of the one-unit wide strip from $x=n$ to $x=n+1$ as small as you want. But if you keep adding up those areas, looking at the whole area from $x=1$ to $x=n$ for bigger and bigger $n$, you find that the even though each new one-unit wide bit adds less to the total than the previous one, the total is unbounded: you can make it as large as you please.
This is the same thing that happens with the harmonic series,
$$\frac11+\frac12+\frac13+\frac14+\ldots\;;$$
the terms get closer and closer to $0$, but the series diverges. That is, you can make the partial sum
$$\frac11+\frac12+\frac13+\frac14+\ldots+\frac1n$$
as large as you please if you take $n$ large enough.
A: Draw the rectangle with base $[1,2]$ and height $\frac{1}{2}$. Then draw a rectangle with base $[2,4]$ and height $\frac{1}{4}$. Then draw a recangle with base $[4,8]$ and height $\frac{1}{8}$. Then draw a rectangle with base $[8,16]$ and height $\frac{1}{16}$. Imagine continuing with that pattern forever.
Note that our first rectangle has area $\frac{1}{2}$. The second rectangle also has area $\frac{1}{2}$. So does the third rectangle, and the fourth, and so on forever. 
Thus by going far enough, we can make the combined area of the rectangles $\gt \frac{1}{2}$, $\gt 1$, $\gt \frac{3}{2}$, and so on forever. So for example by going far enough, we can make the combined area $\gt 100$, $\gt 1000$, and so on. (Of course we will have to go awfully far.)
And now the clincher. All of these rectangles lie between the $x$-axis and the curve $y=\frac{1}{x}$. So if we want, for example, the area under $y=\frac{1}{x}$ and above the $x$-axis to be say $\gt 1000$, all we need to do is to go far enough. There will be an $M$ such that  $\int_1^M \frac{1}{x}\gt 1000$. There will also be a (much larger!) $M$ such that $\int_1^M \frac{1}{x}\,dx\gt 1000000$. And so on. 
