# 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.

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

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.

(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.

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.

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.

8 following

1 answer
1

Are you sure you want to delete this answer?