Математика. Материалы курса

This session focuses on the numerical sequence, or rather on the study of properties of a numerical sequence by convention. All definitions were considered at the lecture and logical entry forms were recorded.

So let's start. The sequence is prescribed by the general term of the formula (see the video.) And there are two objectives: to prove that this sequence is bounded and, secondly, to examine it for monotony.

Let us look at the boundedness. We’ll try to find a few members of this sequence. Calculation (see. Video). What does it mean that sequence is bounded? This means that its members can be included in a segment, generally speaking, calculation of the first few terms does not tell us anything, there is an infinite number of members, they are all different.

Our task is to estimate the fraction from both sides. Obviously, this fraction is greater than zero or equal to zero for any n. Why? For any n, the numerator and denominator are strictly greater than zero Thus, this evaluation is performed for any n.

We have left constraint thus we received low bounded. Then we draw a conclusions. But, what kind of evaluation is on the other hand? You can actually decide differently. It is possible to estimate this fraction using transformations.

We write 2 multiplied by (n + 2) minus 4, all divided by (n + 2). We divide the numerator by the denominator term by term and  get 2 minus 4 divided by (n + 2) (see. Video). Thus, for any n sequence members are obtained by subtracting from 2 a positive number, for every n it will be its own number It is obvious that all these numbers are less than or equal to two. So, what did we get? We can say that this number is two, and this inequation is true for any natural number n small.

Let us revise the definition. We got that all sequence members are between zero and two. So, there were two real numbers - zero and two, that whatever natural n we would take, it would be between these numbers - the sequence is bounded. The first point is completed.

Let's move on to the second point. We need to examine the sequence for monotony. Let us remember the definition. Let’s assume, if the sequence is strictly increasing, then by definition it means that any natural n is strictly less than a with index (n + 1) (if there is decrease, it will be strictly greater, if it is stationary, then there will be equation, if there is increase, the decrease is not strict, then we add an equal sign).

So our task is to find out whether there is a sign which connects и for any n. So let's find. What are we doing? For example, if we were to find, we would substitut n by 3, but we need to find a with index (), then we substitute () instead of n.

We received (2n + 2) divided by (n + 3) (see. Video). We need to find out by which sign и is connected. Let's try to define the sign of the difference, if it is possible for any n, then our task will be completed (see. Video). We write 2n, divided by (n + 2), subtract fraction with the numerator (2n + 2), in the denominator (n + 3). Reduce fractions to a common denominator (n + 2) times by (n + 3) (an additional factor for the first fraction is (n + 3), in the second one - (n + 2)).

Carefully performing calculations we get  minus 4 in the numerator, and product (n + 2) (n + 3) in the denominator. For any natural number n in the denominator we have a plus sign, and in the numerator - minus. So, the fraction turns out less than zero for any arbitrary number n.

We have found that for every natural n a with an index is  less than a with index a (n + 1). We just wrote the definition at the beginning which reflects this case, the sequence is strictly increasing. The problem is solved. Let’s consider the sequence. How can I prove that this sequence is not bounded? What does it mean  a sequence not bounded above?

This means that on the number scale there is not a point that would bound the right side of the sequence (i.e. wherever there is a point, there will always be elements to the right), it will mean unboundedness from above.

Why is the sequence bounded below? It is obvious that for any natural number  n, this expression takes non-negative value (i.e. greater than or equal to zero), that means that the sequence is bounded below. How to write the definition? What are we going to prove?

How will we prove that the sequence is not bounded above? Let us, first, write the definition. If the sequence were bounded above, how would it be written?

So, bounded above means by definition the following - there is such a real number b, that whatever natural number n we would  take, the members of the sequence are less than or equal to b.

Our task is to prove that the sequence is not bounded above. The rule of constructing negation is as follows: each quantifier is reversed (the existential quantifier to a universal one,  the universal quantifier to existential one). Let’s write this definition.

The sequence is not bounded above, if, for any real number b, there exists such index n, that the sequence members with this index will be greater than any preassigned number b (still lower-case n is not fixed here, let us say that here we will look for some index N capital).

So, there is capital N, b with this index N, is over b. We act according to the written definition. Take an arbitrary number b (we do not specify which, it can be 2, 3, -10, - 1.7, it is arbitrary and we reason generally, it does not depend on what is b).

We have to find a formula for N, to set the following program: specify number N for any real number so that the sequence member with that number would be over that b?

Let's first understand, for which indecies N is over b, and if there is any solution. We see that we can take one of the indices N: n + 1 divided by n over b. We are to set a natural number so that the fraction is over b.

Take such a structure: b is an arbitrary real number (it can be negative, we are now trying to construct a natural number that would be over it) and take module b, so we have already hit into a non-negative number.

Let's say thit number is one second - this is not a natural number. Next, using the properties of integeral part, we say that for any real number, there was a number of such type, which is strictly larger. Let's see, what class of numbers it belongs to.

So, the integeral part of a non-negative number is over or equal to zero, we add one, the number is from the set of natural numbers.

Let us denote it N - natural. Obviously, this natural number will be strictly less than the fraction - 1 divided by N, and this is b with index N. We got that for any arbitrary number b, whatever it may be, there was found such natural number N, that b with index N, is strictly larger than b.

So, number N  always exists, in fact, we even found a formula for this number. We got that the sequence is not bounded above. Let us consider another problem. How to prove that the sequence is monotone, we dealt with that, it’s  working with negation again. How to prove that the sequence is not monotone?

What does it mean? This means that the sequence is not increasing, and at the same time is not decreasing. Let's find some elements of this sequence = 0 = 4 = 0 (see. Video). This is enough for us to show that the sequence is not monotone.

We see that the first element is less than the second. This suggests that the sequence can not be decreasing. We do not need to compare all other elements, there is a pair of such two adjacent members that the first is less than the second.

It follows that the sequence is not decreasing.

Let’s consider the following pair of numbers: the second element is larger than the third one (see, with increasing the index sequence members decreased, we started to move to the left along the number scale), so,  the sequence cannot be increasing.