Two reasons I can think of: ⁠It’s easier to represent factorial recursively as n! = n*(n-1)! = n* (n-1)* (n-2)* …. * 2 * 1. ⁠When you’re dealing with factorials in an expression, you often pull out the higher or first few highest terms and that’s easier to do when you represent factorials from largest to smallest.

It's almost transparent. and . You should be able to see it from the above. Or you can argue as follows:

Hint: The factorial of various is the feature that multiplies the number by means of each natural wide variety below it. Hence the concept \\[n!(n + 1)\\] is used in reducing the answer. The concept of factorial is used in permutations and combinations.Complete step-by-step solution:Since we know that factorial or \\[(n!)\\] is the product of all numbers less than that up till \\[1\\]. To get a factorial we have to multiply \\[n\\] with the next number. Hence we use the concept of \\[(n + 1).n!\\] Factorial of a whole range \\[n\\] is defined because it is manufactured from that range with each complete variety until we get a \\[1\\].Hence, we find that that the factorial of \\[n + 1\\] is \\[(n + 1)! = (n + 1).n.(n - 1).(n - 2)...3.2.1\\] Additional Information: The study of factorials is at the primary part of several subjects in mathematics, together with the wide variety of concepts, algebra, geometry, probability, data, graph concept, and discrete arithmetic, etc. The factorial of a number of is the feature that multiplies the quantity via every natural quantity underneath it. Symbolically, factorial may be represented as \"\\[!\\]\" So, \\[n\\] factorial is made of the first \\[n\\] natural numbers and is represented as \\[n!\\] Using the method for factorials we are able to easily derive what's moreover, the definition of the \\[0\\] factorial includes the simplest one permutation of \\[0\\] or no items. Ultimately, the definition also validates some of the identities in combinatorics. Note: One location wherein factorials are usually used is in permutations & mixtures. The factorial function is described for all fine integers, together with 0. Further, in which n = zero, the definition of its factorial (n!) encompasses the product of no numbers, which means that it is equal to the multiplicative identification in broader phrases. The use of factorials is widely done in Permutation and Combination.

By definition, Using the binomial theorem, the term of the development is and

For example,

Just to be clear, as n approaches infinity, 1/n converges to zero. What does not converge however, is infinite series of 1/n. There are several simple and intuitive proofs of this fact. For example, if you group the terms of the series as follows: 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16)...., It is clear that the sum of each group of terms in parentheses is greater than 1/2. We can then keep grouping the terms of the infinite series like this until we get a series which is greater than the infinite series of 1/2, and therefore must diverge. If you are confused why some series with nth terms equal to zero converge and others don't, you can use a simple tool from calculus is the p-series test. A p-series is a series of the form 1/np ,where p is a constant. The p-series test says that if p is less than or equal to 1, the infinite series of 1/np diverges, and if p is greater than 1, the series converges. For the infinite series of 1/n, p=1 and the series must diverge according to the p-series test. Hope this helps.

To show that (n+1)! is in O(n!) you have to show that there is a constant c so that for all big enough n (n > n0) the inequality

(n+1)! < c n!

holds. However since (n+1)! = (n+1) n! this simplifies to

n+1 < c

which clearly does not hold since c is a constant and n can be arbitrarily large.

On the other hand, (n-1)! is in O(n!). The proof is left as an exercise.

For both of these, you can use the rule of n! = n * (n-1)! to rearrange the expressions. Then, you should end up with similar terms in both the numerator and denominator. Using the first example to illustrate what I mean: You can write out (n+1)! as (n+1) * n! , which will turn the entire expression into ( (n+1) * n! ) / n! Since n! is a common factor in both halves of the fraction, you can simplify it by dividing both sides by n!. Dividing anything by itself leaves you with 1, so you'll end up with (n+1) / 1. You can ignore the 1 at the bottom of the fraction, so the whole thing simplifies to n+1. Does this make sense?