Tricki

I have a problem about the convergence of a sequence

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My problem concerns finding the limit of an explicitly given sequence.

• Is the sequence defined by means of a formula for the th term, or is it defined in some less direct way, such as using a recurrence relation?
• It's defined by a formula for the th term.
• Could you say in broad terms what the function for is like?
• is a rational function of , that is, a function of the form where and are both polynomials.
• In that case, all you need to do is to exploit the following facts. If and then , and if It helps to divide through by the largest power of that appears in . For instance, to prove that one rewrites it as and applies the above rules to argue that the top and bottom tend to 1, so the whole fraction tends to 1. It should be noted that this argument freely uses the fact that as This is called the Archimedean property of the real numbers.
• The expression for involves a ratio of two functions that are more complicated than polynomials.
• In that case you have a formula of the form so you want to compare the growth rates of and as tends to infinity. A few techniques can get you quite a long way. For example, if you can find some and prove that the ratio is greater than for every sufficiently large then must tend to infinity. This, for example, is enough to show that tends to infinity. To see this, note that while which tends to as tends to infinity (because it is a product of 100 sequences, all of which tend to ). Therefore, for sufficiently large is bigger than More generally, any exponential function with grows faster than any polynomial function Combining this with techniques for dealing with rational functions will allow you to find limits of sequences such as
• The formula for involves raising a number to a power that depends on .
• There are various things you might have to do here. For instance, if you are asked to prove the obvious seeming fact that as then try turning the problem round: if then ; however, the very simple but surprisingly useful inequality proves that when
If is equal to some number close to raised to a power that depends on , then you will almost certainly want to make use of the fact that For example, if you are asked for the limit of you can argue that it is the cube of and hence that it tends to And is the th root of which is less than (because in fact the sequence is increasing), so it tends to (because tends to ).
• The formula for involves factorials.
• In that case, what you need to do depends on how delicate the problem is. Does it look as though tends to or to ?
• Yes.
• Then you may be able to get away with proving that is bounded away from . For example, if then which implies that Or you may consider using simple estimates for such as that which actually follows from the above calculation if we observe that and
• No (or don't know, or yes but it seems hard to prove).
• In that case the problem is more delicate, and you may need to use Stirling's formula, which says that the ratio of to tends to 1 as tends to infinity.
• It's defined by means of a recurrence.
• There is one trick that is extremely useful in this situation, and it is sometimes even useful for determining limits of sequences where the th term is given by a formula, and that is first to prove that a limit exists, and then to use the recurrence to determine what the limit must be. An example will illustrate the idea. Suppose we define a sequence by taking and If we know that this sequence tends to a limit then both and must tend to (since also tends to ). But tends to , by the rules for adding and dividing limits, so which implies that Since every is positive (by an easy induction), However, we still need to prove that the sequence converges to something. How can we do this without simply proving that it converges to ? The answer is that there are theorems (or axioms) that say that a sequence with such-and-such a property converges. For example, Cauchy sequences converge. For this particular problem, we use the fact that a monotone decreasing function that is bounded below converges. If we know that , then it is easy to check that But and is greater than by the AM-GM inequality applied to and Therefore, the result is proved.

I have an explicitly given sequence, and am required to prove that it converges, but I am not required to calculate the limit.

• In that case, you should consider finding the limit anyway, and you should also consider the techniques for proving convergence of sequences where all you are told is that they satisfy certain properties.

I am supposed to prove rigorously a statement that looks utterly obvious, such as that or tend to

• Such questions look a bit confusing to begin with, but the point is to deduce them from the axioms for a complete ordered field. The simplest technique is the one suggested on this page for dealing with sequences defined by recurrences: argue first that your sequence tends to a limit and then think about what the limit must be. For example, the sequence is monotone decreasing and bounded below by so it converges to some limit (which must be non-negative). If this limit is and then there must be some such that which implies that whenever But this contradicts the fact that tends to A similar argument works for

My problem is to prove that a sequence converges, but rather than being told what the sequence is I just know that it has certain properties.