Cauchy sequences and uniform convergence

[JG89]

Homework Statement

Suppose the infinite series $$\sum a_v$$ is NOT absolutely convergent. Suppose it also has an infinite amount of positive and an infinite amount of negative terms.

Homework Equations

The Attempt at a Solution

Say we want to prove it converges by proving the sequence of partial sums, A_n, are Cauchy.

Then we need to prove that for every positive epsilon, $$|A_n - A_m| < \epsilon$$, for n and m sufficiently large.

Note that $$|A_n - A_m| = |a_1 + a_2 + ... + a_n - (a_1 + a_2 + ... + a_m)| = |a_{n+1} + a_{n+2} + ... + a_m|$$.

Some of these terms are positive, some negative. Are we allowed to "separate" the positive and negative terms? Like this:

Say P_n is the sequence of positive terms and N_n is the sequence of negative terms in the sum $$|a_{n+1} + a_{n+2} + ... + a_m|$$. Then can we write $$|a_{n+1} + a_{n+2} + ... + a_m| = |P_1 + P_2 + ... + P_i + N_1 + N_2 + ... + N_j|$$ ?

It seems to me like this should be fine, since this is a finite sum. But we are going to have to take n and m larger and larger for epsilon getting smaller and smaller, so I am not sure.

 

[Office_Shredder]

What you've given in the question isn't enough to prove the series is convergent, since it still might not be. Can you post the exact question?

 

[JG89]

Alright¸, but my proof is kind of long...

Question: Show that the series $$\sum_{v=1}^{\infty} \frac{cos(vx)}{v^p}, 0 < p < 1$$ for x in the interval [A, 2pi - A], where A is any small positive number.Attempt:

I have shown that the series converges "regularly" by showing that the infinite series $$\sum_{v=1}^{n} cos(vx) \le \frac{1}{|sin(x/2)|},$$ for all n and all non-zero x, and that the sequence 1/n^p is of bounded variation. And so by the Abel-Dedekind-Dirichlet Theorem our original series converges.

Now to prove uniform convergence. Let f_n(x) be the partial sums of our series. I cannot approximate |f_n(x) - f(x)|, where f(x) is our limit function, because I don't know what the series converges to. So I will have to use Cauchy's test:

A sequence is uniformly convergent if and only if for all positive epsilon $$|f_n(x) - f_m(x)| < \epsilon$$ for all x in the interval [A, 2pi - A], provided n and m are taken large enough.

Now, $$|f_n(x) - f_m(x)| = |\frac{cos((n+1)x)}{(n+1)^p} + \frac{cos((n+2)x)}{(n+2)^p} + ... + \frac{cos(mx)}{m^p}|$$.

Some of the terms we are summing are positive, and some of them are negative. This is because of the alternating sign of the cosine term in the numerators. If we were able to make sure that each term in the sum $$|\frac{cos((n+1)x)}{(n+1)^p} + \frac{cos((n+2)x)}{(n+2)^p} + ... + \frac{cos(mx)}{m^p}|$$ was positive, then it would follow that [tex] |\frac{cos((n+1)x)}{(n+1)^p} + \frac{cos((n+2)x)}{(n+2)^p} + ... + \frac{cos(mx)}{m^p}| < $$|\frac{cos((n+1)x)}{(n+1)^p} + \frac{cos((n+1)x)}{(n+2)^p} + ... + \frac{cos(mx)}{(n+1)^p}| \le \frac{1}{(n+1)^p|sin(x/2)|}$$ (I used the fact that the partial sums of the cosine are bounded). Notice that the expression on the most right hand side of inequality goes to 0 for increasing n, regardless of where x is in the interval [A,2pi – A]. And so the sequence is Cauchy for all x in the interval, and so the series converges uniformly.

That was assuming the terms we were adding were positive. So if I could show that all the terms in the sum $$|\frac{cos((n+1)x)}{(n+1)^p} + \frac{cos((n+2)x)}{(n+2)^p} + ... + \frac{cos(mx)}{m^p}|$$ could be rewritten so they are positive, then from what I have posted above, it follows the series converges uniformly. So:

Some terms in our sum are positive and some are negative. Let P_n be a sequence consisting of all the positive terms, and let –Q_n be a sequence consisting of all the negative terms of the sequence. Then $$|\frac{cos((n+1)x)}{(n+1)^p} + \frac{cos((n+2)x)}{(n+2)^p} + ... + \frac{cos(mx)}{m^p}| = |P_1 + P_2 + ... + P_j - Q_1 - Q_2 - ... - Q_i| = |P_1 + P_2 + ... + P_j – (Q_1 + Q_2 + ...+ Q_i|)| \le | P_1 + P_2 + ... + P_j| + | Q_1 + Q_2 + ...+ Q_i|$$.

The rearrangement of the series won’t change the sum because n and m (in the f_n(x) and f_m(x)) are finite numbers, and so there are only a finite amount of terms. Also notice that the Q_n in the very last inequality on the right are positive because we’ve factored out a -1.
From what I posted before, since these terms are now positive, both $$| P_1 + P_2 + ... + P_j|$$ and $$| Q_1 + Q_2 + ...+ Q_i|$$ are bounded by $$\frac{1}{(n+1)^p sin(x/2)}$$ and so $$\sum_{v=n+1}^{m} \frac{cos(vx)}{v^p} \le \frac{2}{(n+1)^p sin(x/2)}$$, and since the right hand side goes to 0 regardless of where x is chosen in the interval [A, 2pi – A], it follows the sequence of partial sums are Cauchy for all x in the interval, and thus the series converges uniformly. QEDEDIT: Where ever you see sin(x/2), it’s suppose to be |sin(x/2)|

 

[JG89]

I just need to know that if we are trying to prove $$A_n = \sum_{v=1}^{n} a_v$$ is Cauchy then can we re-arrange the terms in the sum $$|A_m - A_n| = |a_{n+1} + ... + a_m|$$ since it is a finite sum?

 

[JG89]

Nevermind.

I was able to prove that $$max(| \sum_{v=n+1}^m \frac{cos(vx)}{v^p}|) < 3\epsilon$$ for x in [A, 2pi - A] using Abel's summation by parts formula.