A number-theoretic coincidence

An interesting little tidbit from Tanya Khovanova: consider the Pythagorean triple 3² + 4² = 5², in which all three numbers are consecutive, and another similar coincidental-seeming equation 10² + 11² + 12² = 13² + 14². These are the first two elements of a sequence of similar equations, with k+1 squares on the left-hand side and k squares on the right-hand side, which she presents there.

Still perhaps coincidental is the fact that 3² + 4² = 5² and 3³ + 4³ + 5³ = 6³. Of course once you see these it's tempting to check if 3⁴ + 4⁴ + 5⁴ + 6⁴ = 7⁴. (It doesn't.) In fact, for n ≥ 4,

3ⁿ + 4ⁿ + ... + (n+2)ⁿ < (n+3)ⁿ

and here's a proof. I'll prove it for n ≥ 5. It suffices to show that

(3/(n+3))ⁿ + (4/(n+3))ⁿ + ... + ((n+2)/(n+3))ⁿ < 1.

Call the left-hand side f(n). But the last term on the right-hand side is decreasing in n, and for n ≥ 5, is at most 16807/32768 (its value at 5). Now, we can write the left-hand side of the above inequality as

((n+2)/(n+3))ⁿ (1 + ((n+1)/(n+2))ⁿ + (n/(n+2))ⁿ + ... + (3/(n+2))ⁿ)

and each term in the second factor is at most ((n+1)/(n+2))ⁿ times the previous one. So we have

(1 + ((n+1)/(n+2))ⁿ + (n/(n+2))ⁿ + ... + (3/(n+2))ⁿ) > (1 + r + r² + r³ + ...)

where r = ((n+1)/(n+2))ⁿ. Of course the right-hand side of the previous inequality is 1/(1-r). r is positive and decreasing in n, so 1/(1-r) is as well. Since n ≥ 5, 1/(1-r) is bounded above by its value at 5, which is 16807/9031. Thus for n ≥ 5,

f(n) < (16807/32768)(16807/9031)

and the right-hand side is easily checked to be less than 1.

For n = 4 this proof doesn't work, because I threw away a bit too much in bounding the sum of a finite series by the sum of an infinite one, but it's easy to check.

In fact, f(n) approaches 1/(e-1) as n goes to infinity.

Edit (8:41 pm):: See also The Everything Seminar.