I am stupid and this book is being inconsistent with the Comparison test

[flyingpig]

Homework Statement

Before I state my problem, I need to quote this from my book

http://img18.imageshack.us/img18/3748/comparisonnr.th.jpg

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http://img806.imageshack.us/img806/8665/integraltest.th.jpg

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See the problem here?

Let me draw out the Integral Test for you.

http://img858.imageshack.us/i/droppingcurve.jpg/

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As you can see that for \sum_{n=1}^{\infty } a_n \leq \int_{1}^{\infty} f(x) dx

For even if the integral diverges or converges, the series will diverge or converge respectively. This contradicts the comparison test no?

 

[flyingpig]

Also my professor actually stated this

\sum_{n=1}^{\infty} a_n converges \iff \int_{1}^{\infty} f(x) dx

\sum_{n=1}^{\infty} a_n diverges \iff \int_{1}^{\infty} f(x) dx

Which means if one side is satisfied, the other is immediately satisfied? Is that what if and only if really means when they mean "going both ways"?

 

[SammyS]

(I will refrain from answering the question in the title.) But, there is absolutely NO inconsistency in these. Read the rest of this and think about what each test does and how they're different, before responding with a one or two line post full of ambiguous antecedents (pronouns referring to this or that or it or them or ...).

What does the comparison test compare?

It compares two functions.
Depending on the results of that comparison, it's possible to determine the convergence/divergence of the integral (from a to + infinity) for one of the functions, based upon knowledge of the convergence/divergence of the integral (from a to + infinity) for the other function.​

What does the other test, the integral test, do? More importantly, what two types of things are being compared?

In this test, you have a function and a series constructed in a very special way from a sequence based on that same function. In this case the convergence/divergence is the same for both the infinite series and the integral (from 1 to infinity) of the function.​

So, the comparison test compares the integrals of two functions --- the functions are only related by an inequality. This test does not contain an "if and only if".

On the other hand, the integral test compares a series with the integral of a function. The terms in the series are closely related to the function. This test does have the "if and only if" to which you referred.

The tests are quite different. There is no inconsistency.

 

[flyingpig]

But I am just mainly having problems with the divergence part.

If say f > g

Then by comparison test, if f is divergent, it does not necessarily mean g is. But for Integral test, it says it does (and hence what I mean by the iff)

 

[SammyS]

And a series is not an integral! So what?

 

[flyingpig]

But it can behave like a function

 

[SammyS]

But it can behave like a function

What is it?

I said, "A series is not an integral." Where is a function mentioned in that sentence?

 

[flyingpig]

But it doesn't matter though, the series still decreases so like an inequality between functions.

 

[SammyS]

The series is based on the function in the integral test.

The two functions in the comparison test have little in common.

The two tests are NOT analogous.

 

[Mark44]

Also my professor actually stated this

\sum_{n=1}^{\infty} a_n converges \iff \int_{1}^{\infty} f(x) dx

\sum_{n=1}^{\infty} a_n diverges \iff \int_{1}^{\infty} f(x) dx

Which means if one side is satisfied, the other is immediately satisfied? Is that what if and only if really means when they mean "going both ways"?

I hope he didn't say what you have above. In both statements you are missing important information. In the first statement the right side should also include the the integral converges. In the second statement, it should say that the integral diverges.

 

[Mark44]

But I am just mainly having problems with the divergence part.

If say f > g

Then by comparison test, if f is divergent, it does not necessarily mean g is. But for Integral test, it says it does (and hence what I mean by the iff)

What you're missing here is that the Integral Test doesn't compare two different functions. The function in the integral is the continuous version of the function that is in the series. So for example, if the series happened to be
\sum_{n = 1}^{\infty}a_n = \sum_{n = 1}^{\infty}\frac{1}{n^2}

then the function in the integral would be f(x) = 1/x². Notice that f(1) = a₁, f(2) = a₂, and so on.

 

[flyingpig]

I hope he didn't say what you have above. In both statements you are missing important information. In the first statement the right side should also include the the integral converges. In the second statement, it should say that the integral diverges.

He didn't say that, but that's what he wrote down. Actually someone asked him about the Iff part and he explained it too. That's why I said he "said it"

My notes today were

Thm: For a series \sum_{n=1}^{\infty} a_n satisfying a₁ > a₂ > a₃ > ...> 0, we have

(1) \sum_{n=1}^{\infty} a_n converges \iff \int_{1}^{\infty} f(x) dx converges

(2) \sum_{n=1}^{\infty} a_n diverges \iff \int_{1}^{\infty} f(x) dx diverges

 

[flyingpig]

Also he did a quick proof of this I'll show you

Proof:

(1) \sum_{n=1}^{\infty} a_n, \sum_{n=1}^{\infty} a_n < +\infty

=> \int_{1}^{\infty} f(x) dx < +\infty

so \int_{1}^{\infty} f(x) dx converges

Assume \int_{1}^{\infty} f(x) dx converges, \int_{1}^{\infty} f(x) dx < +\infty

(agree up to this point, but the divergence stuff is flawed I think)

(2) \int_{1}^{\infty} f(x) dx diverges for \sum_{n=1}^{\infty} a_n = +\infty \iff \int_{1}^{\infty} f(x) dx = +\infty

 

[flyingpig]

The series is based on the function in the integral test.

The two functions in the comparison test have little in common.

The two tests are NOT analogous.

But the picture! The picture! I think we only care about the decreasing part of the series/function

 

[flyingpig]

What you're missing here is that the Integral Test doesn't compare two different functions. The function in the integral is the continuous version of the function that is in the series. So for example, if the series happened to be
\sum_{n = 1}^{\infty}a_n = \sum_{n = 1}^{\infty}\frac{1}{n^2}

then the function in the integral would be f(x) = 1/x². Notice that f(1) = a₁, f(2) = a₂, and so on.

Yes I missed that thanks lol

 

[flyingpig]

What you're missing here is that the Integral Test doesn't compare two different functions. The function in the integral is the continuous version of the function that is in the series. So for example, if the series happened to be
\sum_{n = 1}^{\infty}a_n = \sum_{n = 1}^{\infty}\frac{1}{n^2}

then the function in the integral would be f(x) = 1/x². Notice that f(1) = a₁, f(2) = a₂, and so on.

a_n are the values, listing them out? Like what I have in the picture, doesn't change my argument though

 

[SammyS]

Give us a specific example where one of these tests contradicts the other. --- where one says a particular integral or series converges/diverges when that's not true.

 

[flyingpig]

Sammy, I just had an epiphany this morning when I woke up and I figured out why.

The thing is that when we set the series to the function, the series itself is the function, it's just that we take particular n, (namely integers only) and so it behaves as the function itself.

When the comparsion test actually tests perhaps two different functions

 

[Mark44]

Sammy, I just had an epiphany this morning when I woke up and I figured out why.

The thing is that when we set the series to the function, the series itself is the function, it's just that we take particular n, (namely integers only) and so it behaves as the function itself.

That's a pretty awkward way to put it, but that's the basic idea. The function used in the integral test is chosen so that it agees at integer values with the terms in the series.

In the example I used a few posts back, the series to be investigated was
\sum_{n = 1}^{\infty}\frac{1}{n^2}

The function that we would choose would be f(x) = 1/x². The a₁ = f(1), a₂ = f(2), and so on. I said this before, but it sounds like it might make more sense to you now.

When the comparsion test actually tests perhaps two different functions

Not perhaps. It tests two different functions, one of which you know its behavior.