Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

6 Answers 6

In your case, taking $1 \to \infty$:

Series sum over integers $n = 1 \to \infty$: $\displaystyle \sum_{n=1}^\infty f(n)$.

Integrals "sum" over all reals $x = 1 \to \infty$: $\int_{n=1}^\infty f(x) dx$.

Added: this might help in understanding the relationship between partial summation (series) and integration, from Wikipedia's entry on "Series":

Partial summation takes as input a sequence, $\{ a_n \}$, and gives as output another sequence, $\{ S_N \}$. It is thus a unary operation on sequences. Further, this function is linear, and thus is a linear operator on the vector space of sequences, denoted $\Sigma$. The inverse operator is the finite difference operator, $\Delta$. These behave as discrete analogs of integration and differentiation, only for series (functions of a natural number) instead of functions of a real variable. For example, the sequence $\{1, 1, 1, ...\}$ has series $\{1, 2, 3, 4, ...\}$ as its partial summation, which is analogous to the fact that $\int_0^x 1\,dt = x.$

share|improve this answer
Wouldn't the sum of the series be less than the integration? –  Oscar Mar 1 '13 at 4:57
I'm referring specifically to the function (1/2)^x. The integral gives you 0.721 while the series gives you 1. Thanks alot for the help btw –  Oscar Mar 1 '13 at 5:04
$(1/2)^x$: some fractional functions, like the one you mention, with a constant fraction raised to an increasing power: this indeed can be the case... –  amWhy Mar 1 '13 at 5:09
Sorry, it was a typo. The function is (.5)^x –  Oscar Mar 1 '13 at 5:11
:+) ${}{}{}{}{}{}$ –  B. S. Mar 1 '13 at 12:24
show 2 more comments

A series adds together a countable number of terms. An integral "adds" together a function's values over an entire uncountable interval. So $$\sum_{n=1}^\infty f(n),$$ if it converges, adds only contributions from $f|_\mathbb{N}$, while $$\int_1^\infty f(x)dx,$$ if it converges, adds infinitesimal contributions from every real number in $[1,\infty)$.

Here's an interesting note, though. From a more advanced viewpoint, these are two examples of a Lebesgue integral, but with respect to two different measures. The Riemann integral is Lebesgue integration with respect to the Lebesgue measure, which weights all real numbers equally. The sum is Lebesgue integration with respect to a measure supported on the integers, which has the effect of adding up only the values the function takes on $\mathbb{N}$.

share|improve this answer
But if a series is the sum of all (integers?) wouldn't it yield a sum less than the integration? –  Oscar Mar 1 '13 at 4:55
@Oscar Not necessarily. For example, $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ but $\int_1^\infty \frac{dx}{x^2} = 1$. Visualize the sum a left-hand Riemann sum of unit-width blocks approximating the area under $\frac{1}{x^2}$ to see why the sum is greater. –  Neal Mar 1 '13 at 5:48
add comment

More clearly, for series we use $\sum$ which are discrete values summed over integers and for integrals we use $\int_a^b$ which are continuous interval

share|improve this answer
add comment

The integral represents the area under the curve and sums up every value that the curve lies on (not just integer coordinates!). The infinite geometric sum only sums the integer values and is equivalent to an area approximation (a riemann sum).

share|improve this answer
add comment

You can't add together infinitely many anything, buy don't worry because no one can. The series and the integral both represent the limiting behavior of a sum of finitely many terms, i.e.

  1. In the case of a series we are interested in what happens to a finite sum as the number of terms increases without bound.

  2. In the case of the integral we are interested in what happens to a series as a certain factor in each term, called the mesh or the norm of the partition, decreases to zero. Usually this certain factor has a magnitude which depends on the number of terms. For example the factor (b-a)/N if you are doing the Riemann integral of a function over the interval (a,b). "N" in this case is the number of terms in your sum and "1/N" is proportional to the norm of your partition.

This may not be pleasing geometrically but its key to understand what the difference is between a series and a (Riemann) integral. Good question.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.