Range of uniform convergence for a series

[leonardthecow]

Homework Statement

Find the range of uniform convergence for the following series

η(x) = ∑(-1)^n-1/n^x

ζ(x) = ∑1/n^x

with n ranging from n=1 to n=∞ for both

Homework Equations

To be honest I'm stumped with where to begin altogether. In my text, I'm given the criteria for uniform convergence, which is given by

|S(x) - s_n(x)| < ε for all n ≥ N,

and I know that S(x) is defined as

S(x) = lim n→∞ s_n(x).

The Attempt at a Solution

I sort of intuitively grasp what the equations are saying, which is if all the terms cluster together for a large enough n, then the series is said to be uniformly convergent. I am just unsure of how to actually test the given series to see if they meet this criteria, or how to go about assigning values to the criteria listed as the above equations.

Any help would be greatly, greatly appreciated, thanks!

 

[LCKurtz]

What have you tried? Can you find any x where they diverge? Or where they converge? Do you know the Weierstrass M test?

 

[leonardthecow]

Well, the first thought that comes to mind is to use say, the ratio test, to see if the series converge. If I were to do that for η(x), I would end up with (assuming I did the algebra correctly)

lim n→∞ |1/n| = 0,

since the x's in the exponents cancel, which would imply that the series converges for any value of x. From that, I would conclude that the radius of convergence is the set of all real numbers.

However, I'm now unsure of how to use this information (if it is indeed correct) to make a statement about the series' uniform convergence or not.

 

[leonardthecow]

I realized I made a mistake above in applying the ratio test. I think the correct limit should be

lim n→∞ |n^x/(n+1)^x|

In looking at this, for any value of x, it seems that the limit will be equal to 1, which would imply the ratio test is inconclusive.

I have a description of Weierstrass M test in my textbook, but I have difficulty in understanding it and more importantly how to actually apply it to these problems...

 

[LCKurtz]

Think about what you know about $p$ series$$
\sum_{n=1}^\infty \frac 1 {n^p}$$and what that might have to do with your problem.

 

[leonardthecow]

Ahhhhh okay, so if I apply the p-series test, since my starting index is greater than 1, then the series will converge for all x > 1? And the same would be true for ζ(x)?

If that's the case, then I know that the series converge for x > 1, but I'm still left with the problem of determining whether or not the convergence is uniform...

 

[Dick]

Ahhhhh okay, so if I apply the p-series test, since my starting index is greater than 1, then the series will converge for all x > 1? And the same would be true for ζ(x)?

If that's the case, then I know that the series converge for x > 1, but I'm still left with the problem of determining whether or not the convergence is uniform...

That is where the Weierstrass M test comes in. You really have to try harder to understand. Why don't you quote the statement here and try to explain it in your own words. That might help.

 

[leonardthecow]

Well, the first thing I know is that the Weierstrass M test only works in establishing uniform convergence for absolutely convergent series. For ζ(x), the series is clearly absolutely convergent. For η(x), the (-1)^n-1 term will oscillate, so it therefore would be only conditionally convergent (because if the absolute value of the terms were to be taken, the series would diverge (similar to the harmonic series).

Leaving η(x) for a moment then, ζ(x) then can be subjected to the M test.

From the definition, it seems to say that we construct a series such that the nth value of the M series is greater than or equal to the absolute value of the nth term of the series being tested. If M is convergent, then the series under consideration is said to be uniformly convergent.

This seems relatively straightforward enough in theory, but the issue still remains that I've never seen an example of this done (nor had it ever explained in a class), so I'm unsure of how to construct such a series M.

 

[Dick]

Well, the first thing I know is that the Weierstrass M test only works in establishing uniform convergence for absolutely convergent series. For ζ(x), the series is clearly absolutely convergent. For η(x), the (-1)^n-1 term will oscillate, so it therefore would be only conditionally convergent (because if the absolute value of the terms were to be taken, the series would diverge (similar to the harmonic series).

Leaving η(x) for a moment then, ζ(x) then can be subjected to the M test.

From the definition, it seems to say that we construct a series such that the nth value of the M series is greater than or equal to the absolute value of the nth term of the series being tested. If M is convergent, then the series under consideration is said to be uniformly convergent.

This seems relatively straightforward enough in theory, but the issue still remains that I've never seen an example of this done (nor had it ever explained in a class), so I'm unsure of how to construct such a series M.

Ok, that's a reasonable start. The M test is talking about uniform convergence of a series of functions $f_n(x)$, what are the $f_n(x)$ in the case of the zeta function ζ(x)? Could you apply that to show that convergence is uniform on, for example, the interval [2,infinity)?

 

[leonardthecow]

In this case, the series of functions under consideration would just be

f_n(x) = 1/n^x,

So if I were, again using the p-test for this interval, to show convergence on [2,∞) (which is true as long as x > 1), I could say that the nth value of this new series is greater in magnitude than the nth value of the zeta function series by virtue of the fact that the new function's index is offset from the original series' index by 1?

 

[Dick]

In this case, the series of functions under consideration would just be

f_n(x) = 1/n^x,

So if I were, again using the p-test for this interval, to show convergence on [2,∞) (which is true as long as x > 1), I could say that the nth value of this new series is greater in magnitude than the nth value of the zeta function series by virtue of the fact that the new function's index is offset from the original series' index by 1?

I do not understand what you are trying to say here. To prove uniform convergence by the M test you have show that there is a series of constants $M_n$ such the $|f_n(x)| \le M_n$ for all x in [2,infinity). I think you are close, but reread the theorem and try to give me a statement that sounds more like the theorem.

 

[leonardthecow]

So I construct a series

∑M_i

ranging from n=1 to i=∞.

We've established that M_n ≥ |f_n(x)| for all x, and that the M_n series is convergent.

Since the M series converges, there must be a term, call it the Nth term, for which terms afterwards are clustered together within a distance ε.

Since we also know that M_n ≥ |f_n(x)| for all x, there must be an Nth term within this series as well that marks the point at which successive terms will cluster together within a distance ε.

Therefore, by definition, the series is said to be uniformly convergent.

 

[Dick]

So I construct a series

∑M_i

ranging from n=1 to i=∞.

We've established that M_n ≥ |f_n(x)| for all x, and that the M_n series is convergent.

Since the M series converges, there must be a term, call it the Nth term, for which terms afterwards are clustered together within a distance ε.

Since we also know that M_n ≥ |f_n(x)| for all x, there must be an Nth term within this series as well that marks the point at which successive terms will cluster together within a distance ε.

Therefore, by definition, the series is said to be uniformly convergent.

No, you still aren't thinking about this right. What concretely is $M_n$? How do you know $|f_n(x)| \le M_n$? Don't worry about ε. The M test will take care of that part.

 

[leonardthecow]

M_n is just a number, correct? And (maybe this is my point of non understanding then) we're assuming for the sake of the test that |f_n(x)| ≤ M_n, so the previous statement is true by definition?

I realize that logic is more or less circular, where I'm assuming what I'm trying to prove in order to prove it...I really just do not see how to explicitly create such a series of numbers. I see that 1/n^x will converge for a certain range of x, but beyond that, it's still very unclear to me.

 

[Dick]

M_n is just a number, correct? And (maybe this is my point of non understanding then) we're assuming for the sake of the test that |f_n(x)| ≤ M_n, so the previous statement is true by definition?

I realize that logic is more or less circular, where I'm assuming what I'm trying to prove in order to prove it...I really just do not see how to explicitly create such a series of numbers. I see that 1/n^x will converge for a certain range of x, but beyond that, it's still very unclear to me.

Yes, you are being circular. Exactly right. Earlier you correctly said $f_n(x)=1/n^x$. What is an upper bound for $|f_n(x)|$ on the range x in [2,infinity)?

 

[leonardthecow]

Well, the function decreases as n gets larger, so the largest value it can take will be 1/2^x, assuming x is positive

 

[Dick]

Well, the function decreases as n gets larger, so the largest value it can take will be 1/2^x, assuming x is positive

No, no. FIX n. What's the maximum as a function of x where x is an element of [2,infinity)?

 

[leonardthecow]

Ah I'm sorry, I misread that. Well the maximum for a fixed n would then be 1/n², since the value of the function would decrease as the denominator gets larger with x

 

[Dick]

Ah I'm sorry, I misread that. Well the maximum for a fixed n would then be 1/n², since the value of the function would decrease as the denominator gets larger with x

Exactly, now we are cooking. So $|f_n(x)| \le M_n$ where $M_n=1/n^2$. So what does the M test tell you about convergence of that series of functions on [2,infinity)?

 

[leonardthecow]

The convergence of that series of functions would have to be less than or equal to 1/n² for x in the interval [2,∞), which is true as shown by the result of the p-test since the p-test shows that the series converges for x>1, which intersects the set [2,∞)...right?

 

[Dick]

The convergence of that series of functions would have to be less than or equal to 1/n² for x in the interval [2,∞), which is true as shown by the result of the p-test since the p-test shows that the series converges for x>1, which intersects the set [2,∞)...right?

You still aren't saying what I want you to say. Would you quote the statement of the Weirerstrass M-test exactly and compare with what you just said? I was really hoping you would say something about uniform convergence on [2,infinity).

 

[leonardthecow]

Here goes...

Let ∑f_n(x) from n=1 to n=∞ be a series of functions all defined for a set of values of x. If there is a convergent series of constants ∑M_n from n=1 to n=∞, such that |f_n(x)| ≤ M_n for all x in the set of values of x, then the series is uniformly convergent in the set of values of x.

What we have is M_n=1/n² for the set of x values [2,∞), with a maximum value of course equal to 1/n² and which converges by the p-test.

Additionally, we have f_n(x)=1/n^x, which converges for x>1.

So, by the M test, there is uniform convergence on the interval [2,∞), since the series for M_n converges in the interval, and since the magnitude of f_n(x)=M_n.

If this is correct, could the statement be generalized to accommodate an interval 1+ε, where ε >0? Which would give a final answer that shows uniform convergence on the interval 1+ε≤x<∞?

 

[Dick]

Here goes...

Let ∑f_n(x) from n=1 to n=∞ be a series of functions all defined for a set of values of x. If there is a convergent series of constants ∑M_n from n=1 to n=∞, such that |f_n(x)| ≤ M_n for all x in the set of values of x, then the series is uniformly convergent in the set of values of x.

What we have is M_n=1/n² for the set of x values [2,∞), with a maximum value of course equal to 1/n² and which converges by the p-test.

Additionally, we have f_n(x)=1/n^x, which converges for x>1.

So, by the M test, there is uniform convergence on the interval [2,∞), since the series for M_n converges in the interval, and since the magnitude of f_n(x)=M_n.

If this is correct, could the statement be generalized to accommodate an interval 1+ε, where ε >0? Which would give a final answer that shows uniform convergence on the interval 1+ε≤x<∞?

Yes. You have it exactly correct. That's how you use the M-test.

 

[leonardthecow]

Wow that a struggle...I really appreciate all the guidance and persistence though, it definitely helped more than the textbook or lecture did. Thank you!

 

[Ian Baughman]

I am way late to this thread but I had a question on η(x). The problem provides the answer of uniform convergence being in the set of 0 < s ≤ x < ∞. I was able to prove that it does exhibit uniform convergence on the interval of 1 < s ≤ x < ∞ but how do you start by showing that it does down to zero?