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.
Update: 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!
Thus,
(1+n)^n = ∑[k=0 to n] C(n,k)/n^k < ∑[k=0 to ∞] 1/k! = e
Update 2: 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 +... show more 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.
Update 3: 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... show more 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.
Update 4: 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.
Update 5: 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... show more 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.
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