Placing Expressions in Increasing Order

[rayroshi]

On page 332 of Mark Ryan's book, Calculus for Dummies, he discusses a useful tip to try, before launching into an elaborate test to determine whether an infinite series diverges or converges. The tip is to just compare the numerator and the denominator of the expression in question; if the numerator is "larger" than the denominator, then the series diverges, otherwise, it converges.

And to make the tip quick and easy to use for purposes of comparison, when inspecting the numerator and denominator, he lists four expressions listing them from "smallest" to "biggest," as follows: n^10, 10^n, n!, and n^n.

This is where my question arises. When I plug in a number for n in those expressions, they don't fall in the same order as he states; for example, if I plug 5 into "n," then I get the following: n! = 5! = 120 ; n^n = 5^5 = 3,125 ; 10^n = 10^5 = 100,000, and ; n^10 = 5^10 = 9,765,625. In other words, my list, in ascending order, would be n!, n^n, 10^n, n^10, which is different than his list.

Am I missing something here? Am I not understanding what he means? Or is he wrong?

Any help would be much appreciated.

 

[HallsofIvy]

On page 332 of Mark Ryan's book, Calculus for Dummies, he discusses a useful tip to try, before launching into an elaborate test to determine whether an infinite series diverges or converges. The tip is to just compare the numerator and the denominator of the expression in question; if the numerator is "larger" than the denominator, then the series diverges, otherwise, it converges.

I presume that the word "larger" is in quotes because he is using it in a very special situation. For example, the sum \(\displaystyle \sum \frac{1}{n}\) has denominator, n, always larger than the numerator, 1, but does not converge.

And to make the tip quick and easy to use for purposes of comparison, when inspecting the numerator and denominator, he lists four expressions listing them from "smallest" to "biggest," as follows: n^10, 10^n, n!, and n^n.

So this is the special sense in which he is using "larger".

This is where my question arises. When I plug in a number for n in those expressions, they don't fall in the same order as he states; for example, if I plug 5 into "n," then I get the following: n! = 5! = 120 ; n^n = 5^5 = 3,125 ; 10^n = 10^5 = 100,000, and ; n^10 = 5^10 = 9,765,625. In other words, my list, in ascending order, would be n!, n^n, 10^n, n^10, which is different than his list.

Am I missing something here? Am I not understanding what he means? Or is he wrong?

Any help would be much appreciated.

The point is NOT that one is larger than the other for some specific number but that if we go out far enough on each sequence, it will eventually be larger.

 

[Deleted member 4993]

If you check n>25 - you'll see that the suggested order - by the book - holds good there after.

 

[rayroshi]

I presume that the word "larger" is in quotes because he is using it in a very special situation. For example, the sum \(\displaystyle \sum \frac{1}{n}\) has denominator, n, always larger than the numerator, 1, but does not converge.


So this is the special sense in which he is using "larger". Sorry about the late response. Five years is a little slow, I guess.


The point is NOT that one is larger than the other for some specific number but that if we go out far enough on each sequence, it will eventually be larger.

Ah! Okay, then it all makes sense. Thanks for that reply.