Problem: Investigate convergence of the following series. (Does it converge? If it does, then how?)

Solution: We have to check on convergence of the given series and also on the absolute convergence of it. We start with the former. What can we say about the terms of this series? It is well-known that sin(k) attains positive and negative values in a totally irregular way, moreover those values are never exactly 1 or −1, the function sin(x) simply picks numbers from the range (−1,1), usually it takes three positive and then three negative ones, but sometimes, just for variety and without any pattern, it throws in four positive numbers or four negative numbers. In particular, there is no way to use the Alternating series test and the typical calculus course offers no tool to handle such series.

Here in Math Tutor we went a bit further, some advanced courses also mention less popular but sometimes powerful tests, in particular we introduced the Dirichlet test. Can it be used in our situation? The numbers bk = 1/k definitely satisfy the assumptions there, so all that is left is to check that the terms ak = sin(k) form a series whose partial sums are bounded.

Actually, this is a very good question. We already recalled that we usually have three positive and then three negative and then three positive numbers etc. going on, occasionally four of like signs come, so it seems feasible that they would nicely keep cancelling. This is indeed true, but proving it is another matter. A relatively simple argument can be made using complex numbers and a couple of well-known formulas, in particular the formula for partial sums of geometric series.

We have shown that all partial sums are bounded by a common constant, therefore the given series satisfies assumptions of Dirichlet's test and as such it is convergent.

By the way, the number in the denominator above is 2sin(1/2).

Some people may be put off by those complex numbers above and ask: Since the question has nothing to do with complex numbers, is there a way to prove this boundedness just using real numbers? Actually, there is, but it is at least as tricky as the complex numbers approach above. It starts with this observation. One particular trig identity can be used to express sin(k) as follows:

Why on earth would we do such a thing? Notice that we transformed sin(k) into a difference and we will be adding many such terms, which should ring a bell: telescopic series.

Uff. I do not know of a nice proof. By the way, we obtained the same upper bound as with the complex numbers approach.

 

Now we pass to investigation of the absolute convergence of the given series. We are supposed to decide convergence of the series

What test will help us? The sine suggests that we try comparison. We get

Unfortunately, the series on the right is the harmonic series whose divergence is well-known, for instance by the p-test. Thus this comparison does not help at all. Note that the sine in absolute value picks values from the whole range (0,1), therefore for some k the term |sin(k)|/k is basically zero, while for other k it is pretty much 1/k. In other words, the estimate we used is as tight as possible, we cannot go lower with the upper bound.

Since |sin(k)| does not have a limit at infinity, it makes no sense to claim that it looks like this or that for k large, so there is no way to use limit comparison.

The lack of convergence of the sine part also rules out limit versions of the Root test and the Ratio test. Since we obviously cannot integrate the sine in absolute value (and the relevant function is not non-increasing anyway), we cannot use the Integral test either and we just ran out of tests to use. Therefore it is definitely worth our time to return to the Root test and Ratio test and inquire about the more general inequality versions of them.

We know from experience that the k in the denominator does not help at all, since in these two tests it gives 1. We can see it when we attempt to use the limit version.

Thus if we want to establish some inequality, we have to use the sine part for it. And here we are totally out of luck. The sine part is never more than 1, so we cannot prove divergence in the Root test. On the other hand, sometimes sin(k) is so close to 1 that [sin(k)]1/k gets arbitrarily close to 1 as well and we cannot separate it from 1 as needed for convergence. Thus also the inequality version of the Root test does not help.

We run into similar trouble with the Ratio test. The ratio sin(k + 1)/sin(k) can be almost 0, but it can be also very large, definitely more than 1. Thus it is not possible to establish any inequality needed in that test.

We just ran out of tests that people usually know, and also the less popular tests from section More tests in Theory - Testing convergence do not offer any way out.

We are forced to conclude that the tests that we know and standard approaches are not good enough to determine convergence of this series; this means that so far we do not know whether the given series converges absolutely.

Note that if we did not have the Dirichlet test, we would have known absolutely nothing about our series. The Alternating series test could not be used; in such a case we typically pass to absolute convergence, but here we also ran afoul of troubles. So the tools that are traditionally covered in calculus courses are powerless in this problem. There actually is a way to decide convergence, but it requires that we look closer at the given series.

It is essentially the harmonic series whose terms were modified. What happens to them? We know that for infinitely many k the numbers |sin(k)| are essentially 1, so many terms of the harmonic series (that is known to diverge) are actually preserved in our series. On the other hand, we know that sin(k) is essentially 0 for infinitely many k, so great many terms of the harmonic series drop out when we modify it. Thus convergence or divergence of our series depends on how many terms get (almost) preserved compared to those that (almost) disappear. In other words, we need to know more about what is happening in the sine function.

A good start is observation that for every small value of |sin(k)| there also must be a rather large value, thus overall there are more "large" values than small values. Indeed, imagine the graph of sine. If we want |sin(k)| to be almost zero, then the integer k must fall almost next to some intersection of the sine graph and the x axes; that is, k must be essentially nπ for some integer n. But then the next integer, k + 1, must fall near the top of the sine wave and as such it must yield (after being substituted into sine) a relatively large number. The same also works for k − 1, so we indeed have much more large numbers than "almost zeros" among those |sin(k)|. This suggests divergence (we preserve a large part of the harmonic series when modifying it), but are not there yet. We did argue that "there is more large terms than almost zeroes," but this does not allow us to say that, say, at least half of the terms in the series we investigate are large, because this pairing that we investigated only concerns extreme values, while majority of terms in the series are actually somewhere in the middle - and we do not know anything about these yet. If we are to succeed, we have to find out something about them.

There are at least two approaches to do this, two possibilities to get a more tangible observation. First, note that the couple k, k + 1 has the interesting property that the corresponding values get balanced no matter where it is positioned with respect to the sine waves. Just take a segment of length 1 (segment [x,x + 1] for x real) and slide it along the x-axis in the picture.

It seems that at least one end always yields large value of sine, sometimes both. Since we do not know which ends is the large one, the best way to express this is to actually talk about the sum of the two values. We claim that there is a certain positive number a such that the sum of |sin(x)| and |sin(x + 1)| is always at least a. This can be shown using standard methods, in fact this magical lower bound is sin(1), for details see this note.

Having this proved we can show that our series diverges. First we group its terms into couples.

Since all the terms in this series are positive, the associative law can be applied here, (this is actually not so obvious as it seems, see Basic properties in Theory - Introduction). Thus convergence of the series on the left is equivalent to convergence of the new series on the right. However, there we can use comparison.

This proves that the series on the left is divergent, so the series with absolute values is divergent and therefore the given series is not absolutely convergent.
Conclusion: The given series converges conditionally.

Alternative: When imagining what can happen on the graph of sine we may notice also this: When we take three successive integers, then at least one of them makes |sin(k)| large, namely greater than 1/2. This is actually proved quite easily, see this note.

Having this established we proceed similarly as before. First we group our series into triplets.

Now we look at one such triplet, for simplicity we put some positive constants on the top and using our observation we know that at least one of them is greater than 1/2. Thus we can estimate

Using this in the decomposition of our series into triplets we obtain

This confirms divergence of our series.

So we won in the end, but it was tough work. Note that there are series that beat even the most advanced tests and tricks, for instance convergence or divergence of the following series, although it is rather nice, is still unknown.

(At least it was in 2004 when I last checked, so even if it were not true now, it shows that a simple series like this can withstand several centuries of attacks by all kinds of convergence tests and tricks.)


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