Calculus II please explain integral test

[iScience]

i understand the reason for if the evaluated integral converges then the series also converges. However, just because the evaluated integral diverges, why does this automatically mean that the related series also diverges?

the integral consists of every number of the +x axis when evaluating f(x), the function's convergence or divergence to be determined. so if the sum of the f(x) at every single x point/interval on the +x axis is evaluated to converge, then this obviously means that the series would converge because the series consists of much less f(x) values to be summed up, even though the sequences for both the integral and the series function follow the same trend, ie converge to the same value.

However.. for the integral test, when there exists a function f(x) whose evaluated integral turns out to diverge, why does this automatically mean that the series (whose sum consists of much less f(x) values..) also diverges?

 

[Millennial]

The integral test follows from a theorem in mathematics called the Euler-Maclaurin summation formula. The exact statement of the theorem is this:

"Let $f(x)$ be continious, positive and decreasing on the interval $[1,\infty)$. Then, the sequence defined by $\displaystyle a_n = \sum_{k=1}^{n} f(k) - \int^{n}_{1} f(x) dx$ is decreasing and bounded below by zero. Hence, it converges."

What this theorem implies is that if the integral has a finite value, the series also must have a finite value; and if the integral is divergent, then so must be the series (because their difference is always finite.)

Does that clarify the situation?

 

[Stephen Tashi]

so if the sum of the f(x) at every single x point/interval on the +x axis is evaluated to converge, then this obviously means that the series would converge because the series consists of much less f(x) values to be summed up,

If that were valid reasoning then we wouldn't need the hypothesis that the function is nonnegative and monotone decreasing. There are analogies between integrals and sums, but an integral isn't a summation of f(x) over every single point in the interval of integration.

However.. for the integral test, when there exists a function f(x) whose evaluated integral turns out to diverge, why does this automatically mean that the series (whose sum consists of much less f(x) values..) also diverges?

An intuitive way to look at it is to graph the graph of a nonnegative monotone decreasing function $f(x)$ and represent a Riemann sum drawn under the graph by a drawing rectangles with base $[n-1,n]$and height $f(n)$ at each integer $n$. Then graph a shifted version of $f(x)$ as $g(x) = f(x+1)$. The area under this graph is less than the area of the rectangles. The improper integrals $\int_0^{\infty} f(x) dx$ and $\int_0^{\infty} g(x) dx$ will both converge or both diverge since they are related by a simple change of variable. So if $\int_0^{\infty} f(x) dx$ diverges, then $\int_0^{\infty} g(x) dx$ diverges and the area under$\int_0^{\infty} g(x) dx$ is less than the sum of the area of the rectangles, which is the sum of the series $\sum_{n=0}^\infty f(n)$.

 

[iScience]

@ Millennial

wouldn't the equation you wrote (btw how do you write it in such a nice format? :D ) be negative? that's what i kinda meant in my question. because the integral on the right contains every value in the epsilon sum on the left, but NOT every value in the integral is contained in the summation on the left, therefore the integral would contain more values, so if you subtract the integral from the summation, this value would be negative, no?

 

[Millennial]

You are wrong, because the sum gives a so-called right sum, greater than the integral. To see this for yourself, graph the curve $y=x$ on the Cartesian plane. Now, draw rectangles in each interval $[x,x+1)$ starting from $x=0$ with height $x+1$. The sum of all those rectangular areas is the value of the summation. Now tell me, which one is greater? The sum or the integral?

 

[Stephen Tashi]

how do you write it in such a nice format?

This thread describe how to do it: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

On the one hand, using LaTex in posts can be frustrating - but if you've ever done computer programming, its a familiar kind of frustration (dealing with syntax errors). You'll want to use the "Preview Post" function of the editor a lot before submitting the final version of a post.

On the other hand, the ability to use LaTex is useful in many places besides this particular forum. You'll find that different math forums use different "tags" to delimit the LaTex. The Wikipedia uses "<math>" and "</math>" instead of the square bracketed "tex" and "itex" tags that are used on this forum. Some forums use the "tex" tag but not the "itex" tag.

 

[Mark44]

On the other hand, the ability to use LaTex is useful in many places besides this particular forum. You'll find that different math forums use different "tags" to delimit the LaTex. The Wikipedia uses "<math>" and "</math>" instead of the square bracketed "tex" and "itex" tags that are used on this forum. Some forums use the "tex" tag but not the "itex" tag.

A fairly recent addition here at PF is to delimit the LaTeX stuff with $$ (same as [tex]) or ## (same as [itex]).

 

[Millennial]

A fairly recent addition here at PF is to delimit the LaTeX stuff with $$ (same as [tex]) or ## (same as [itex]).

I didn't know that, testing:
$$e^{ix} = \cos(x) + i\sin(x)$$

Seems to work fine. Nice!