Prove (1+1/n)^(n+1/2)>e for all positive integers n?
Prove (1+1/n)^(n+1/2)>e for all positive integers n.
Preference given to answers that have the least
amount (preferably none) of integration/differentiation.
Here's an example proof of a slightly different problem (that (1+1/n)^n<e):
Clearly n! < (n^k) * (n-k)! for n ≥ k > 1. Therefore
n!/(n^k * (n-k)!) < 1 so dividing by k! means
C(n,k)/n^k < 1/k!
(1+n)^n = ∑[k=0 to n] C(n,k)/n^k < ∑[k=0 to ∞] 1/k! = e
Nice job, Scythian. But could you finish the proof off? First, I agree that you've shown
(1 + 1/(n(n+2)) )^(2n+2) > (1+2/n) > 0 because the first 3 (3rd powers are not needed) terms on the left are:
1 + C(2n+2,1)/(n²+2n)+C(2n+2,2)/(n²+2n)² =
1 + 2(n+1)/(n²+2n) + ½*2(n+1)(2n+1)/(n²+2n)² >
1 + 2(n+1)/(n²+2n) + 2/(n²+2n) =
1 + 2(n+2)/(n²+2n) = 1 + 2/n
Thus, you've shown:
((n+1)² / (n(n+2)))^(2n+2) > (n+2)/n or
(1+1/n)^(2n+2) > (n+2)^(2n+3) / (n * (n+1)^(2n+2)) or
(1+1/n)^(2n+1) > (1+1/(n+1))^(2n+3) and square rooting yields:
(1+1/n)^(n+½) > (1+1/(n+1))^(n+3/2).
In other words, (1+1/n)^(n+½) is monotonically decreasing.
But how do you conclude that (1+1/n)^(n+½) > e? Haven't you only shown that you have a decreasing sequence (bounded below by 1)? Also, I missed where the induction comes into play.
Define T(n,a) = (1+1/n)^n * √(1+1/(n+a)).
You showed T(n,0) is strictly monotonic decreasing (and bounded below by 1). I showed that T(n,∞) < e (but I did not show that the bound is tight). And it is easy to see that T(n,0) > T(m,∞) for all integer n,m. However, this only guarantees that T(n,0) converges to a number less than or equal to e.
This question is really asking whether either the T(n,0) or T(n,∞) bound can be made tight (one implies the other) without using calculus, and using the infinite sum as the definition of e.
By the way, you are correct in needing 3rd powers for showing (1 + 1/(n(n+2)) )^(2n+2) > (1+2/n). The ">" is incorrect in what I wrote.
Ah, we're using different definitions for e - I was using the series. So the argument is something like: T(n,0) is monotonic decreasing and since all the terms in its product are at least 1 it is clearly bounded below by 1. Hence, T(n,0) approaches some value.
T(n,0)>T(m,∞) for all positive n,m. Since T(n,0)-T(n,∞) becomes arbitrarily small there is a common value that T(n,0) and T(n,∞) converge to, and we call this value e = lim [as n -> ∞ of] (1+1/n)^n. Fine, that answers the question.
I may post a followup question to this as there are a few topics in here that strike me as interesting. For example, showing that T(n,a) is monotonic increasing for integer a>0. And showing that the series definition (that I was using) gives the value for e.
I'll leave this open a little longer in case you have any follow on thoughts.
- Scythian1950Lv 71 decade agoFavourite answer
It can be done without calculus. We prove by induction, but it does get slightly messy. We know that for n = 1, (1+1/1)^(3/2) = 2.82843 > e. We wish to show that (1+1/n)^(n+1/2) > (1+1/(n+1))^(n+3/2) for any n > 1. After strenuous rearrangment (knowing that n is always positive), we end up with, to show:
(1 + 1/(n(n+2)) )^(2n+2) - (1+2/n) > 0
For n > 1, we expand this expression and leave out all powers of 1/(n(n+2)) greater than 3 (which we can do because they are all positive). What's left, we simplify and end up with:
(n² +3n + 8) / 3n²(n+2)³ > 0
which is always true for n > 1
Edit: Oh, that's right, an oversight. We know that in the limiting case n -> ∞, the expression reduces to e (or at least > e). If somehow it ends up being LESS than e before n reaches ∞, then at some point in time, it would have to climb back up again. But this would contradict the induction proof that every successive term would decrease, and not ever increase.
Edit: I don't understand by what you mean by T(n,0) is "bounded below by 1", and "only guarantees that T(n,0) converges to a number less than or equal to e"? It's bounded below by e, because should it fall below e for any n, it has to increase at some point in order to reach the limiting value of e. We know that (1+1/n)^n = e for n = infinity, and we know that (1+1/n) is not < 1. And we can show that for any n, T(n+1,0) > T(n,0).
Edit 2: While it's true that for a < 0.15, T(n,a) > e and for a > 0.19 T(n,a) < e, for values 0.15 < a < 0.19 the picture gets messy, i.e., T(n,a) equals e for some finite n. That's to be expected, since otherwise there would exist a value for a such that T(n,a) = e for all n.