Does the series converge using the integral test?

[opus]

Homework Statement

Use the integral test to compare the series to an appropriate improper integral, then use a comparison test to show the integral converges or diverges and conclude whether the initial series converges or diverges.

$\sum_{n=3}^\infty \frac{n^2+3}{n^{5/2}+n^2+n+1}$

Homework Equations

The Attempt at a Solution

I'm really just at a loss on what to do here. I went to the tutor lab on campus and three tutors (one of which was the class TA) couldn't answer this question.
So to start somewhere, I can compare this series to an integral via the integral test. However, to use this test, I would need to evaluate the integral and to my understanding the integrand of the integral would be of the form $f(n)=a_n$ for all integer values n greater than or equal to 3. My first problem is that there is no way I can evaluate $\int_3^\infty \frac{x^3+3}{x^{5/2}+x^2+x+1}dx$. So there must be something else that I can do, but I'm not sure what. Any ideas?

 

[BvU]

Hi,

no way I can evaluate $\ \int_3^\infty \frac{x^3+3}{x^{5/2}+x^2+x+1}dx$

You should be more interested in $\int_3^\infty \frac{x^2+3}{x^{5/2}+x^2+x+1}dx\$ $\ \$ which is still uncomfortable I suppose .

However, with limits it's a good idea to divide numerator and denominator by a power that makes things more manageable, in this case $x^2$. Does that help ?

 

[opus]

Hi,
You should be more interested in $\int_3^\infty \frac{x^2+3}{x^{5/2}+x^2+x+1}dx\$ $\ \$ which is still uncomfortable I suppose .

However, with limits it's a good idea to divide numerator and denominator by a power that makes things more manageable, in this case $x^2$. Does that help ?

Why $x^2$ in particular?

 

[dRic2]

The integral is (there should be $x^2$ in the numerator instead of $x^3$... just to be sure)
$$\int_3^\infty \frac{x^2+3}{x^{5/2}+x^2+x+1}dx$$
The lower limit is a nice finite number and doesn't bother us, the real problem is to understand the behaviour of the function at $\infty$. Of course the function goes to zero at $\infty$ but will the area under the function go to zero as well ? For example $\frac 1 x \rightarrow 0$ but sadly $\int_1^{\infty} \frac 1 x = \infty$. On the contrary $\frac 1 {x^2} \rightarrow 0$ and $\int_1^{\infty} \frac 1 {x^2} = 1$. This means that $\frac 1 {x^2}$ goes to zero "fast enough", while $\frac 1 x$ is slow. What can you say now, about your function ? How does it behave at $\infty$ ?

 

[BvU]

Why $x^2$ in particular?

Did you learn about limits in your curriculum ?
So that towards $+\infty$ you have a good idea of the behaviour of the integrand

 

[Ray Vickson]

Homework Statement

Use the integral test to compare the series to an appropriate improper integral, then use a comparison test to show the integral converges or diverges and conclude whether the initial series converges or diverges.

$\sum_{n=3}^\infty \frac{n^2+3}{n^{5/2}+n^2+n+1}$

Homework Equations

The Attempt at a Solution

I'm really just at a loss on what to do here. I went to the tutor lab on campus and three tutors (one of which was the class TA) couldn't answer this question.
So to start somewhere, I can compare this series to an integral via the integral test. However, to use this test, I would need to evaluate the integral and to my understanding the integrand of the integral would be of the form $f(n)=a_n$ for all integer values n greater than or equal to 3. My first problem is that there is no way I can evaluate $\int_3^\infty \frac{x^3+3}{x^{5/2}+x^2+x+1}dx$. So there must be something else that I can do, but I'm not sure what. Any ideas?

The title of your post implies that the integral test is necessary, but the body of your post suggests that what is wanted is a convergence/divergence decision about the series, and that the integral test is something you decided to use.

I think the integral test is just about the hardest way to do the problem. If $T(n) = (n^2+3)/(n^{2.5}+n^2+n+1)$ is the $n$th term of your series, why not try to find a simpler form $f(n)$ such that $0 < T(n) < f(n)$ or $T(n) > f(n) > 0$, where the convergence or divergence of $\sum f(n)$ is easy to assess? The task can sometimes be simplified a lot by noting that for some fixed, finite $N$ the first $N$ terms of the series won't affect the ultimate convergence or divergence of the series, so it is enough to have a "comparison" $f(n)$ that is valid for $n > N$. In your case, getting an appropriate $f(n)$ is very easy.

 

[opus]

Thanks all. Since I don't quite follow your arguments, I've had to go back and go over the prior material on this stuff because I feel like there's some key aspect that I've missed. I'll have a better response in a couple of hours when I've reviewed the material. Thanks so much for your replies.

 

[Delta2]

You can find another integral that it is a lower bound for that cumbersome integral. For example you can do

$\frac{x^2}{x^{5/2}+x^{5/2}+x^{5/2}+x^{5/2}}\leq \frac{x^2+3}{x^{5/2}+x^2+x+1}$

and I believe you can prove that the integral $\int_3^{\infty}\frac{x^2}{4x^{5/2}}dx$ diverges hence your original integral diverges too.

 

[opus]

Why can we use other integrals such as that? To make the point of my confusion clear, let me state the definition of the integral text as listed in my text:

Suppose $\sum_{n=1}^\infty a_{n}$ is a series with positive terms $a_n$.
Suppose there exists a function $f$ and a positive integer $N$ such that the following three conditions are satisfied:
1) $f$ is continuous
2) $f$ is decreasing
3) $f(n)=a_n$ for all integers $n \geq N$

Then $\sum_{n=1}^\infty a_n$ and $\int_S^\infty f(x)dx$ both converge or diverge.

Now where I am not following here is #3 in the definition. When choosing another integral, like you have done, how can you be sure that $f(n)=a_n$ for your new integral?

 

[Delta2]

3) won't be satisfied for the new integral. The new integral is introduced just to help us decide whether the original integral converges or not, the new integral is not meant to satisfy the integral test conditions. With the help of the new integral we just use a comparison test for the original integral, to show that it diverges that's just it.

 

[dRic2]

Why can we use other integrals such as that?

if $f(x) \le g(x)$ almost everywhere then $\int_E f(x) dx \le \int_E g(x) dx$

The idea id to find a simple integral that is easy to compute. If such integral is zero then $\int_E f(x) dx$ has to be zero for the property I mentioned above.

 

[opus]

Ok so that makes more sense now. But forgive me for being a little nitpicky, but say we have three functions- the original $f$, $g$, and $h$.
Now say our $f$ is similar to the one in our problem, and we want to use it for an integral test for the series, but it's a little unruly to integrate. So we choose a function that is similar to $f$, which we will call $g$ and we have $g$ greater than or equal to $f$. Now say we evaluate $g$, and find that it is divergent, and from this, we conclude that if it is divergent, then $f$ is divergent as well since $f$ is less than $g$. But, what if there exists a function $h$ such that $f \leq h \leq g$ that is convergent? In other words, we didn't pick our new integral to be close enough to get an accurate representation of the end behavior of $f$?

 

[Delta2]

First of all you , by using the comparison criterion for integrals, you need $g<f$ and if g diverges then f diverges too. Now this also means that if you find h such that $g<h<f$ h will also diverge by the same comparison test. In other words, trust me, if you have found g and undoubtedly concluded that g diverges, then every h such that $g<h$ diverges too.

f and g don't have to be close enough, it is sufficient for the criterion that $g\leq f$. It doesn't matter how big or small the difference $g-f$ will be.

 

[opus]

Ok this has eliminated a lot of my confusion and it's not quite as mysterious as I had originally thought. Thank you!

 

[Ray Vickson]

Why can we use other integrals such as that? To make the point of my confusion clear, let me state the definition of the integral text as listed in my text:

Suppose $\sum_{n=1}^\infty a_{n}$ is a series with positive terms $a_n$.
Suppose there exists a function $f$ and a positive integer $N$ such that the following three conditions are satisfied:
1) $f$ is continuous
2) $f$ is decreasing
3) $f(n)=a_n$ for all integers $n \geq N$

Then $\sum_{n=1}^\infty a_n$ and $\int_S^\infty f(x)dx$ both converge or diverge.

Now where I am not following here is #3 in the definition. When choosing another integral, like you have done, how can you be sure that $f(n)=a_n$ for your new integral?

Now that others have given the solution, I can explain my original post, which claimed that the integral test was not needed. The point is that if your summed terms are $T(n)= (n^2+3)/{(n^{2.5}+n^2+n+1)}$, then the numerator is $>n^2$, and for $n > 1$ the denominator is $< n^{2.5}+n^{2.5} + n^{2.5} + n^{2.5},$ so
$$\sum_n T(n) > \sum_n \frac{n^2}{4 n^{2.5}} = \frac{1}{4} \sum_n \frac{1}{\sqrt{n}} = \infty.$$ Basically, converting $\sum T(n)$ into $\int T(x) \, dx$ is a waste of time if the integral $\int T(x) \,dx$ is not easily doable. Bounding the integrand (as done in #8) is just the same as bounding $T(n)$, so there is no reason at all to integrate. The integral test is very useful if the integral is easy to do, or is some well-known value that you can look up somewhere, but otherwise I doubt its use as a first choice of method.

 

[opus]

Now that others have given the solution, I can explain my original post, which claimed that the integral test was not needed. The point is that if your summed terms are $T(n)= (n^2+3)/{(n^{2.5}+n^2+n+1)}$, then the numerator is $>n^2$, and for $n > 1$ the denominator is $< n^{2.5}+n^{2.5} + n^{2.5} + n^{2.5},$ so
$$\sum_n T(n) > \sum_n \frac{n^2}{4 n^{2.5}} = \frac{1}{4} \sum_n \frac{1}{\sqrt{n}} = \infty.$$ Basically, converting $\sum T(n)$ into $\int T(x) \, dx$ is a waste of time if the integral $\int T(x) \,dx$ is not easily doable. Bounding the integrand (as done in #8) is just the same as bounding $T(n)$, so there is no reason at all to integrate. The integral test is very useful if the integral is easy to do, or is some well-known value that you can look up somewhere, but otherwise I doubt its use as a first choice of method.

Awesome thank you! Looks like these types of problems require some experience and good judgement.

 

[LCKurtz]

I just got here so I'm going to add my 2 cents worth. I would have just observed that the convergence of the series, which is the quotient of polynomial-like terms, is determined by the highest order terms in the numerator and denominator. That immediately suggests using the general comparison test with the series whose n'th term is $$\frac{n^2}{n^{2.5}} = \frac 1 {\sqrt n}$$Of course, this is essentially the comparison others have suggested, but you don't have to worry about whether the numerator or denominator is less or greater than what.