[bonanova]

Caveat: not for the faint of heart.

Several have noted that for an infinite series to converge, the sequence must converge to zero.

That is, the terms of the series must have zero as a limit. For example the series

1/2 + 1/4 + 1/8 + 1/16 + ... + = 1

would not converge if the terms 1/2ⁿ did not approach 0 as n-> infinity.

For those familiar with the integral calculus, where the integral can be treated as the limit

of a series, a corollary seems to be that, for the improper integral of a function to exist

[converge], the function, if it's continuous, must have a limit of 0.

So we ask: is that true?

That is, if

.

1. f(x) is defined, continuous, and differentiable for 0<= x <= infinity, and
2. Integral [x=0, infinity] f(x) dx exists and is finite,
   

.

then must f(x) have a limit as x-> infinity, and must that limit be zero?

Clearly if it's not true then there must be a ... counter example.

[Guest]

quote name='bonanova' date='31 July 2009 - 12:10 PM' timestamp='1249035043' post='191053']

f(x)is defined, continuous, and differentiable for 0<= x <= infinity

Edited July 31, 2009 by DeeGee

[Guest]

sin(x)

On second thoughts isnt the intrgral between 0 and infinity undefined for this function?

Or does that not matter?

Edited July 31, 2009 by psychic_mind

[Guest]

I don't quite understand the question. Testing for convergence is not really that simple from what I remember. There were at least a couple of different ways to test for convergence and each one is different from the other. The ones that I remember were the Comparison Test, Ratio Test, Root Test, Integral Test, Limit Comparison Test, Alternating Series Test, and for uniform convergence there's the Weierstrass M-Test. So either I'm completely off in what I'm thinking about or I don't get the question.

Please explain.

[Guest]

Well, the initial assumptions were that

# f(x) is defined, continuous, and differentiable for 0<= x <= infinity, and

# Integral [x=0, infinity] f(x) dx exists and is finite

So, sin(x) and ln(x) don't really work for this, I don't think.

I think that the statement given is true, though, and we can prove it. By the way, I'm going to also say that in order for an integral to exist, it must be finite, anyway.

So let's say that there exists some function, f(x), that is defined, continuous, and differentiable (I don't think that the differentiability matters here, but...) on [0,inf). Additionally, let's suppose that this function's limit as x tends to infinity is not zero. We can approximate this integral by using a Riemann sum, like so...

Let's say that the width of all of the rectangles is 1, so the approximation of the integral would be just A = A₁ + A₂ + A₃ + ...

Clearly, the actual integral, Integral [x=0, infinity] f(x) dx > A, since the rectangles are completely underneath the curve. Since the limit as x tends to infinity of f(x) is not zero, then the limit as n tends to infinty of A_n is not zero either. Furthermore, this means, from one of the original assumptions, that A = A₁ + A₂ + A₃ + ... cannot converge. Since the integral is greater than A, it cannot exist either.

Q.E.D.

Edited July 31, 2009 by Chuck

[Guest]

Clearly if it's not true then there must be a ... counter example.

sin(x)/sinh(x)?

[Guest]

Here's the three basic rules that we can all agree on. 1) If a series is convergent then the xth term gets closer and closer to 0 as x gets closer to infinity. 2) If the xth terms does not get closer and closer to 0 as x get closer to infinity then it's divergent. 3) There may exist a divergent series where the nth term gets closer to 0 as x gets closer to infinity. What you're asking for is an example of the third occurrence. Good examples of #3 is 1/x and 1/log(x). In both cases they are divergent but in both cases the sum approaches 0. Just very slowly. But this brings forth another complication. Absolute convergence. 1/x is divergent. But ((-1)^x)/x is convergent. However it is not an absolute convergence. This is why I was a bit confused in the beginning. Testing for convergence is a pain in the a**. For example ((-1)^n/sqrt(n)) is convergent but is not an absolute convergence.

[EventHorizon]

If f(x) does have a limit that is not 0, then we know the integral cannot converge to a finite value. But does a limit need to exist at all? I say no.

So we need a function that is continuous and differentiable and will integrate to some finite number, and that you cannot choose any small value z such that after some point |f(x)| < z.

Start with cos(x). I will use the part of it from a peak to a valley. Now add straight lines from where the peak and valley were. This gives the main unit of my function. It gives a continuous and differentiable way to move the function from one constant value to another. I will now just say what the various heights of the function are, without referring to how to keep the function continuous and differentiable. I will also not worry about the integral of this area, since it would be trivial to accommodate.

let f(x) start at -1 until the integral reaches a value of W.

f(x) would transition to a value of 1 until the integral reaches 0.

f(x) would then transition to -1/2 until the integral reaches a value of W/2.

f(x) would then transition to a value of 1 until the integral reaches a value of 0.

The process would repeat with f(x) taking these constant values over certain ranges of x:

{-1,1,-1/2,1,-1/4,1,-1/8,1,-1/16,1,...}

The stretches of f(x)=1 would decrease in length by about half every time, but the stretches of negative values would stay essentially the same length, but halving in height each time. f(x) will keep moving from 1 to negative values and back, so no limit exists. It is continuous and differentiable though. If you look at the integral, it looks like a bunch of sawteeth that halve in height as x->infinity. You can choose any value e>0 you want, and there is a value of x for which the integral of f from 0 to any value greater than x will be less than e. So the integral converges, but the function itself has no limit.

So, did I overlook anything? *ducks and covers head*

[Guest]

the harmonic series which is 1/n converges to 0 but its summation is still infinity. doing the integral test in reverse would say the same must be true for 1/x. And it is. the integral of 1/x from 0 to infinity is ln of infinity minus ln of 0. ln of 0 is basically -infinty so the answer is basically 2*infinity and therefore infinity.

[bonanova]

I received this answer as an email from a new member Mathephobe who had trouble posting it.

It arrived about 2:00 pm EDT July 31 2009.

Dear Bonanova, I'm just a new member - as of today - but I wanted to contribute to your posting of earlier today.

But there is a simple counterexample to the statement that f(x) --> 0 as x-->inf if the improper integral of f from 0 to inf is finite.

Here's how it goes:

Create a function f, so that f(x) = 1 whenever x = n, a positive integer, and f(x) = 0 elsewhere

(this function is not continuous, but we will modify it a bit so it is even smooth, i.e. differentiable).

Now complete the construction of f by constructing a thin little smooth bump on the graph of f around x = n,

say over the interval [n - 1/2^n, n + 1/2^n], for each n = 1,2,3,...

(e.g. when n = 3, say, the interval is [3 - 1/8, 3 + 1/8])

and where f is 0 at the end points of the interval.

Note the area under such a bump is < 1/2^(n - 1),

hence the 'area' under such an f , from 0 to inf, is < 1 + 1/2 + 1/4 + .... = 2.

QED

Good problem that I hadn't run into before.

And congrats on your impressive history on Brainden.

Yours, Mathephobe

[Guest]

I still don't get it though. How is my answer wrong?

[bonanova]

I still don't get it though. How is my answer wrong?

Can we come up with a function f(x) with these properties for 0 <= x < = inf?

.

1. f(x) is defined, continuous and infinitely differentiable [no piecewise linear portions]
2. f(x) > 0 [strictly positive]
   
3. Integral [x=0, inf] exists and is finite.
4. But still it is not the case that lim [x->inf] f(x) = 0.

[Guest]

Got it. Thank you for clearing it up. I see it now

[EventHorizon]

Can we come up with a function f(x) with these properties for 0 <= x < = inf?

.

1. f(x) is defined, continuous and infinitely differentiable [no piecewise linear portions]
2. f(x) > 0 [strictly positive]
   
3. Integral [x=0, inf] exists and is finite.
4. But still it is not the case that lim [x->inf] f(x) = 0.

It's not pretty, but...

f(x) = lim a->infinity, sum (b=0 to a), 1/(1+ ( (2^b) * (x-b) )^2 )

For a given b, the equation is essentially 1/(1+x^2) squished horizontally by 2^b then moved to the right by b.

1/(1+x^2) integrated from -infinity to infinity is pi. Squishing horizontally by 2^b reduces the area under the curve by dividing by 2^b. Integrating the above from -infinity to infinity would result in the summation pi(1+1/2+1/4+1/8+...), which sums to 2*pi. So the integral from 0 to inf exists and is finite.

f(x) for any value of x can by calculated as the sum of an infinite series (which will always converge).

f(x) is continuous and infinitely differentiable

f(x) is strictly positive

Integral [x=0, inf] exists and is finite. (less than 2*pi since -inf to inf is 2*pi and f(x)>0 for all x)

f(x) is greater than 1 for every integral value of x >= 0, and f(x) dips down to near 0 (but always positive) between integral values. Therefore, lim x->inf f(x) doesn't exist.

[bonanova]

Mathephobe and EventHorizon have solved it. Kudos.

Emulate a convergent infinite series by placing smooth functions at integer values of x.

The functions have a specified height [say of 1], so their sum does not have a limit of zero at infinity.

Individually they have an integral on the infinite domain that is less than the terms of the convergent series.

The rest is straightforward.

Modify the infinitely differentiable Gaussian function to emulate terms of the series 1 + 1/2 + 1/4 + 1/8 + ... + = 1.

Analysis:

Since Integral _-inf ^inf exp {-ax²} dx = pi/a we can simply let a = 2ⁿ.

Then Integral _-inf ^inf exp {-2ⁿx²} dx = pi/2ⁿ. We'll construct our function from these terms.

Let f_n(x) = exp {-2ⁿ(x-n)²}. Note that f_n(x) = 1 when x = n instead of when x = 0, but its integral is still pi/2ⁿ.

Finally, define f(x) = Sum _n=1 ^inf f_n(x).

Because f(x) > 1 at integral values of x, f(x) does not have a limit of 0.

Then Integral ₀ ^inf f(x) dx < Integral _-inf ^inf f(x) = pi Sum _n=1 ^inf 1/2ⁿ = pi. So the integral exists.

The strict inequality over the smaller integration range owes to the fact that f(x) > 0 everywhere.

Edit: There's a square root missing in the above analysis - an easy fix.

But I'm tired, and it doesn't invalidate the result.

I was asked this question in an oral exam.

Edited August 5, 2009 by bonanova
Point out error and add personal note