Problems with the area of the graph of a function

Note that the answer has been changed.

Then construct a series a[n], with a[n] representing the length of the line segment that the owner was talking about when x=n. Then write a generalized formula, similar to the way you learned in high school to find S[n] (sum of first n terms) from the generalized formula for a series. As long as the interval you give is finite, even if S[n] is hard to express, you can do it the dumb way one by one, and you can always figure it out. So this problem is still much simpler than the area problem.

For example, if 1/x is taken when x only takes 1, 2, or 3, wouldn't that be 1/1+1/2+1/3=11/6? If x=1,2,...n, then the sum of the lengths is S[n]=1+1/2+1/3+...+1/n, which can't be written as a generalized expression.

But the infinite case is troublesome and requires the use of level theory. For example, you say x=1,2,3 ...... infinite go down this infinite number of line segments total length is what? It's the fact that the number of steps 1+1/2+1/3+...... keeps adding up, and it's divergent, that is, n tends to infinity when S[n] has no limit, it's an infinite amount. But some functions do. For example, say y=1/x?, you say x takes the sum of all the lines going down 1, 2, 3 ...... infinity, and S[n]=1+1/2?+1/3?......+1/n?, which takes the limit as n tends to infinity, is equal to π?/6 (as to why Inexplicably appear pi π, this is not a word can not be clear, this conclusion is also more difficult to prove).

The owner asked the wrong question, right?1+1/2+1/3...+1/nI've already said that it can converge to infinity, and it can't be less than 2. What the owner said should be 1+1/2?..+1/3?.........+1/n?...<2This is a typical deflationary method. The topic of proving inequalities should have been covered by your high school teacher. All 1/n?<1/n(n-1)=1/(n-1)-1/n is then 1+1/2?+1/3?......+1/n?<1+(1-1/2)+1/2-1/3)+1/3-1/4)...+(1/(n-1) -1/n)=2-1/n<2. and it is proved.