If $f$ is positive, decreasing, and continuous on $[0,+\infty)$ show that $\int_0^\infty f(t)dt$ converges iff $\sum_{n=0}^\infty f(n)$ converges My Work:
Integrals take infinitely small steps to get from one term to the next, whereas in a series, the distance between each term must be some tangible value ($|x_n-x_{n-1}|\ge \epsilon$). Thus, in a decreasing positive and continuous function, the value of the series will always be greater than the value of the integral. Therefore, the integral will converge if and only if the series converges by the comparison principle.
My solution seems very basic, and not holistic. I do not feel that the answer is correct. If anyone could assist me in finding a solution that is more in depth, or show me the error in my logic, I would greatly appreciate it
 A: There are two major pieces to this proof. I'll leave you to figure out how to use them (and how to show them.)
First, for any positive function $f$:


*

*$\sum_{n=0}^\infty f(n)$ converges if and only if $g(m)=\sum_{n=0}^{m} f(n)$ is bounded.

*$\int_{0}^\infty f(x)\,dx$ converges if and only of $g(x)=\int_{0}^x f(x)\,dx$ is bounded for $x\geq 0$.


Second, if $f$ is decreasing:
$$f(n)\leq \int_{n-1}^n f(x)\,dx \leq f(n-1)$$
A: For each positive integer, n, consider the rectangle with vertical sides at n-1 and n, base y= 0 and top y= f(n).  You should be able to see that
(1) the area of that rectangle is f(n) and 
(2) For every n, the top of that rectangle is below the graph of y= f(x) because f is decreasing.
  So the sum of the areas of all those rectangles is less than the integral of f.
Now consider the rectangle with vertical sides at n and n+ 1, base y= 0 and top y= f(n).  You should be able to see that
(1) the area of the rectangle is f(n) and 
(2) for every n the top of that rectangle I above the graph of y= f(x).
