Sure, there are many ways.  Alternating series can be pretty easy.  For example:

I already gave you a major hint that it is an alternating series.

But in general, what you try to do is take the first few summations, and come up with a sequence.  Then you must prove that it actually is the sequence through induction (which normally isn't too bad).  Then you just have to show what number that sequence converges to, which can be very painful or even just about impossible.

I didn't really read through your proofs, but you got the answer right.  But there is an easier way to do it.  As I said, this is an alternating series, though it doesn't look like it.

Now when you write the sum, it will look like:

But we can't reorder terms until we prove that the series is convergent.  So let's prove the series is convergent:

And now we're done.  We can regroup terms, and so the sum is 1, since all the fractions cancle themselves out.

Fourier series are so difficult.
By the way, do you know what are cosh(x) and sinh(x) used for, Zhylliolom ?

You need a good book on real analysis.  In my experience, infinite sums are either really easy, or almost impossibly hard.

Ricky: I feel a little silly now that I didn't approach it that way. But I arrived at the same answer, so I suppose it isn't too bad. I like your method better though.

Great!  Now you know how I felt when I had my advanced calculus final a while back with this exact problem on it, and I was the last one in the room with only this problem to go.

Edit: One little thing about your proof though, you want the base case to be from i=1 to 1.  Otherwise, it is very clear and well written.

Edit #2: I also didn't see your question at the top.  An alternating series is simply a series that... alternates, with signs that is.  So the first term is positive, the second term is negative, the third positive, then negatve, and so on.

It most certainly isn't standard.  How in the world did you come up with that?

Great news, folks. After a haunting dream and a good portion of the evening spent sitting furiously with my notebook and pencil, I have found a simple equation which can be used to calculate ζ(2n). No need for Fourier series or wacky factoring approaches, this formula works for all positive integers n. After finding this equation, an immediate consequence was noted: it could be used to find ζ(-n) as well, and also I will show how this miraculous formula can be used to prove a fundamental property of the zeros of the Riemann zeta function(no skipping ahead to see this part, let the suspense build).

Let's start by defining Bernoulli numbers. The Bernoulli numbers are the coefficients B_n of this series expansion:

There isn't really a simpler way to define them, aside from a definition using a contour integral. The first few Bernoulli numbers are

From those few values you can probably notice one thing: for integers n > 0, B_{2n + 1} = 0, or in other words, odd values of n greater than 1 for B_n give the value zero. This concept will be important much later in this exposition, so try to remember it. Of course, you will be reminded again anyway.

Now

isn't really in a form we can play with much for our present situation. Let's convert it into something we know more about. We can add x/2 to it and get something decent:

So to summarize,

Now we try to work something out for coth using our knowledge of B_n's generating function:

Now we already established that for all odd positive integers n > 1, B_n = 0. Thus we can make the substitution n = 2k in the above sum given that we make up for the loss of the term corresponding to n = 1. A quick calculation shows that the term we need is equal to -x/2, so we must add this in for our substitution n = 2k to work:

Alright, great. Now we would prefer something more familiar than the hyperbolic cotangent to work with. Let's use the substitution x = 2iz to get our standard circular cotangent(note that coth ix = -i cot x):

Beautiful series there. But this is no time for sightseeing, we have an equation to derive. Let us consider the known identity(whose proof is left to the reader as an exercise )

Let's clean up the sum a little to make it more workable. Start by making the substitution x = z/π and multiplying through by it:

Now let's push our manipulation of this even farther:

An observant reader may notice that our summand is actually the value of the geometric series

Because of this fact, we will substitute this sum in for the summand and upon manipulation make a marvelous discovery:

So, after all that work, we now have two sums for z cot z, and we may write

Let's kick z cot z out of the picture now, as we have used him as much as we need to:

Now we want to get rid of the summations. It turns out that such a task is easy. We want the limits of summation to be equal, so let's get the lower limit on the left sum to be k = 1. We can do this by simply writing the lower limit as k = 1 and adding the value corresponding to k = 0 to the left side. The corresponding value is 1, so we now write

Now just subtract 1 from each side:

The summands must be equal now:

Solve for ζ(2k) to get

I prefer the expression with the imaginary unit, but some may be more comfortable rewriting it as

So there it is, a nice little formula for ζ(2n). But that was only part of what will be discussed in this post. Let's take things a step farther and find a formula for ζ(-n) as well! To do this we'll use our newly discovered formula for ζ(2n) and the functional equation for the zeta function,

Make the substitution s = 2n:

Use our formula for ζ(2n) and simplify:

Letting k = 2n we get the form

Now substituting n = k - 1, we get

Now we have the power to evaluate the zeta function for negative integers! Here's a few values:

We can also plug in n = 0 to get

Now, as promised, I will prove a fundamental property of the zeros of the Riemann zeta function. The Riemann hypothesis states that for s ≠ -2, -4, -6... such that ζ(s) = 0, Re(s) = 1/2. Let's use our nice equations to figure out why the hypothesis is stated the way it is, and to prove something about ζ(s) = 0.

The statement of the Riemann hypothesis disallows values of s that are negative even numbers. Why is this? Using the derived equation for ζ(-n), we can see that for negative even integers, the Riemann zeta function always seems to have a zero. These zeros must be kept out of the hypothesis because they obviously have real part not equal to 1/2. But one may ask this question about ζ(-2n): "Is it always zero?" The answer is yes, and I shall prove it for you.

From our formula for ζ(-n), we can write ζ(-2n) as

A long while ago, I told you that for all integers n > 0, B_{2n + 1} = 0. So if that is true, ζ(-2n)'s denominator will always be zero and thus ζ(-2n) will always be zero. So our task now is to prove that B_{2n + 1} = 0 for all integers n > 0.

Recall the expression

from earlier. We know that B₁ is an odd value of n in B_n which does not equal zero, so we take its term out of the series and write a little note by our sigma:

Now give x a negative value in the equation and note that coth(-x) = -coth x:

It turns out that x coth x is an even function, and since it is equivalent to the sum, the sum must also be an even function. Thus we require that (-1)ⁿB_n = B_n, which only holds for even n, or in another way of stating the situation, all odd powers of x in the sum must have a coefficient of 0. Thus all B_{2n + 1} = 0 for integers n > 0 and ζ(-2n) is always zero.

Last edited by Zhylliolom (2006-08-13 08:10:31)

Just need a bit of help with your variables.  k, n are natural, x is real, and z is complex, right?

To

Is the identity supposed to be:

Ok, I admit I rushed the post a little(I was excited to share this with everyone), so a few things might be confusing. In my notes, I used z and s (Riemann style) for the variables. Consider x and z to be both complex. Should I change x to s in my original post?

Ricky wrote:

To

Is the identity supposed to be:

No, I did a few steps in one there(probably the only part of my proof where I made a small jump, I think). I substituted x = z/π to get this:

Then I multiplied through by x = z/π to get

Next, z²/π² was moved inside the sum, and inside the sum we clear fractions by multiplying by π²/π²:

So I think the part that confused you there was me getting from π cot πx to z cot z, but as you can see, the substitution cleared the π and gave us a z inside the cot, and the multiplication of both sides by z/π put a z outside the cot and cancelled out the π outside, giving us z cot z.

Last edited by Zhylliolom (2006-08-10 10:24:59)

Zhylliolom, or maybe someone else could explain what happened here, I'm a bit puzzled :

I noticed the geometric sum after that, but I've never worked with double sums before, ever. But does it basically mean that you take the sum over the sum, as in (k=1 to ∞) first and then (n=1 to ∞) over that previous sum?

If so, I learned something new by reading your post

Zhylliolom wrote:
/*But this is no time for sightseeing, we have an equation to derive. Let us consider the known identity(whose proof is left to the reader as an exercise wink)

*/
I give up, would you give me a hint?