TITLE: Calculus: Improper integrals vs series
QUESTION [9 upvotes]: What is the difference between improper integrals and the a series?
For example, if you solve a type one improper integral from 1 to infinity, the answer is different than if you solve the same function using a geometric series.

REPLY [3 votes]: I don't know but it sounds like the OP may be asking about the Euler-Maclaurin Formula. The Euler-Maclaurin formula describes exactly what is the "difference" between an infinite sum and an improper Riemann integral. Let us consider the function $f(x)=\frac{1}{x^2}$ for example on the domain $[1,\infty).$ You may know that $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges and $\int_1^{\infty}\frac{1}{x^2}dx$ converges.
This comes from the integral test which tells us the improper integral converges if and only if the infinite sum converges (assuming whatever conditions). But the values of the sum and the integral can be different. In our case the sum is $\frac{\pi^2}{6}\approx1.64493...$ and the integral is exactly 1. You can use the convergence of the sum/integral to conclude the convergence of the other. But you can't use the value of one to conclude the value of the other.
So then the question is well how close are the values to each other. I mean they can't be "too" far off because the sum is just a discrete version of the integral and they are both converging and have roughly the same shape. The answer is that the Euler-Maclaurin Formula tells us exactly how the VALUES of the sum and the integral are related to one another. It describes the difference between the sum and the integral by using the higher derivatives of $f$ and the Bernoulli numbers.