Using the notation we show it.
It is easy to use Napierís inequality to show that
Raise e to those powers .
Raise each of those to the nth power
It is squeezed between e & e.
I'm reviewing some basic calculus concepts, and I have the following questions that will probably be very easy for you guys to answer, although I'm stuck since a long time on them.
Given the usual limit that is used in the definition of "e": lim (n->infinity) (1+1/n), how do we prove that the limit exists and it is finite?
Can we use the binomial theorem expansion for a finite n, and prove that the coefficients tend to be 1/k! for n that goes to infinity, so we get the Taylor expansion? I'd like to see a rigorous proof of this derivation of e's expression as a power series, we shouldn't forget to prove that the coefficients of the binomial expansion don't diverge (in fact, i guess they should converge) as n goes to infinity...
Or without expanding it, can we prove the following statements:
1-prove that a(n) is strictly increasing
2-prove that a(n) is limited above by a finite positive number L (for instance, we could try to prove it for L=3 or L=4): a(n)<L for all n.
A-if we manage to prove the two statements above, then did we successfully prove that the limit exists?
B-is there a better way than this to prove that the limit exists?
Thanks! And please excuse me for the length of the post...
Thank you for answering so soon, the proof is very simple, but how do you define ln(x) without proving the limit of e first? I mean, wouldn't it be circular logic?
We should find a way to prove it without using the natural logarithm...
Alternatively, we could define ln(x) as the integral of 1/t from 1 to x, and then napier's inequality should be simple to infer, but we need to use a integral definition...
there should be an elementary method to prove it...
Another useful way to define ln(x) is simply:
Since is defined for all non-zero t, this defines ln(x) for all positive x. It is clearly continuous and differentiable for all x. By the fundamental theorem of Calculus, its derivative is 1/x. Since that is positive for all positive x, ln(x) is an increasing function and so a one-to-one function.
It is not difficulty to prove, by substitutions in the integral, that
1) For any positive x, .
2) For any positive x and y, .
3) For any positive x and y any real number, .
For example, for the last one, is, by this definition, .
If , define so that , so that . When t= 1, and when , so the integral becomes
(If y= 0 then and still.)
It follows then that ln(x) is a one-to-one function from the positive real numbers onto the real numbers and so has an inverse function, exp(x), from the real numbers to the positive real numbers.
Finally, if y= exp(x), then x= ln(y). If x is not 0, divide both sides by x to get . Going back to the exponential form, so .
(If x= 0, then y= 1 because . So we still have .