# Deciding if Divergent/Convergent via Comparison Theorem

• drofenaz
In summary, the improper integral of (1/x.5) is divergent. When you calculate its integral within a closed interval [0,a] you get a finite answer. However, the improper integral (regarding the startpoint of the interval [0,b]) is convergent, as weird as it may seem.
drofenaz

## Homework Statement

Use the Comparison Theorem to determine whether the integral is convergent or divergent.
$\int_{0}^{1}\frac{e^{-x}}{\sqrt{x}}$

## The Attempt at a Solution

$\int_{0}^{1}\frac{e^{-x}}{\sqrt{x}} \leq \int_{0}^{1}\frac{1}{x^{\frac{1}{2}}}$

Because p<1, the integral is divergent. Wolfram, however, is giving me a finite answer, meaning the integral is actually convergent. Where am I going wrong?

Last edited:
The improper integral of (1/x.5) is divergent. When you calculate its integral within a closed interval [0,a] you get a finite answer.

You're actually doing a reasonable comparison because (e-x/x.5) will (for x≥0, real numbers) always be under (1/x.5), therefore, the integral between 0 and a (in this case, a=1) will always be less then the other, since it's the defined area between the graphs and the x axis.

Hence, you can conclude it's convergent.

Mathoholic! said:
The improper integral of (1/x.5) is divergent. When you calculate its integral within a closed interval [0,a] you get a finite answer.

You're actually doing a reasonable comparison because (e-x/x.5) will (for x≥0, real numbers) always be under (1/x.5), therefore, the integral between 0 and a (in this case, a=1) will always be less then the other, since it's the defined area between the graphs and the x axis.

Hence, you can conclude it's convergent.

I'm not sure I quite understand. When the comparison equation is near 0, the graph is up at +∞. So how is it possible that it is convergent?

The improper integral (as the endpoint of the interval [0,b] approaches ∞) is divergent. Meaning that if you calculate the area below the graph in this interval [0,∞[ you don't get a finite answer.
However the improper integral (regarding the startpoint of the interval [0,b]) is convergent, as weird as it may seem, because we're talking about the sqrt(x)-1. If instead of having that you had x-1 (they both have alike graphs), the same improper integral would give ∞ at x→0.

It's really interesting to see that expressions with basically the same graph have different responses in what concerns limits/integrals.

Mathoholic! said:
The improper integral (as the endpoint of the interval [0,b] approaches ∞) is divergent. Meaning that if you calculate the area below the graph in this interval [0,∞[ you don't get a finite answer.
However the improper integral (regarding the startpoint of the interval [0,b]) is convergent, as weird as it may seem, because we're talking about the sqrt(x)-1. If instead of having that you had x-1 (they both have alike graphs), the same improper integral would give ∞ at x→0.

It's really interesting to see that expressions with basically the same graph have different responses in what concerns limits/integrals.

So it's because of the limits that it is convergent. If it were from 1 to 0 it would be divergent?

drofenaz said:
So it's because of the limits that it is convergent. If it were from 1 to 0 it would be divergent?
No. Changing the limits of integration just changes the sign of the value of the integral.

Assuming that f is integrable on [a, b],

$$\int_a^b f(x)dx = -\int_b^a f(x)dx$$

I was mixing up my p-test for 0->∞ and 0->1.

## What is the Comparison Theorem in science?

The Comparison Theorem is a mathematical concept used to determine whether a series or sequence converges or diverges. It states that if a series or sequence can be shown to be smaller or larger than a known series or sequence that converges or diverges, then the original series or sequence will have the same behavior.

## How is the Comparison Theorem used to decide if a series or sequence is divergent or convergent?

The Comparison Theorem is used by comparing the original series or sequence to another known series or sequence that either converges or diverges. If the original series or sequence is smaller or larger than the known series or sequence, then it will have the same behavior, indicating whether it is divergent or convergent.

## What is the difference between a divergent and convergent series or sequence?

A divergent series or sequence is one that has no finite limit and its terms do not approach any specific value, while a convergent series or sequence is one that has a finite limit and its terms approach a specific value as the series or sequence progresses.

## Can the Comparison Theorem be used for all series and sequences?

No, the Comparison Theorem can only be used for series and sequences that have positive terms. It also cannot be used for alternating or oscillating series or sequences.

## Are there any limitations to using the Comparison Theorem?

Yes, the Comparison Theorem only gives an indication of whether a series or sequence is divergent or convergent. It does not provide an exact value for the limit of a convergent series or sequence. Additionally, it can only be used for certain types of series and sequences, as mentioned before.

Replies
5
Views
2K
Replies
3
Views
571
Replies
4
Views
614
Replies
2
Views
691
Replies
2
Views
1K
Replies
3
Views
1K
Replies
5
Views
2K
Replies
2
Views
1K
Replies
6
Views
787
Replies
1
Views
912