TITLE: Proving $(2n-1)^n + (2n)^n ≈ (2n+1)^n$
QUESTION [10 upvotes]: As I do, I was messing around and I thought to myself this simple thing:
$3^2 + 4^2 = 5^2$
I just thought that this is only Pythagorean triplet with sequential integers. I know that there are no others and there are no others to higher powers due to Fermat's Last Theorem. However there are many that can be approximated.
$5^3 + 6^3 = 341 ≈ 7^3$
$7^4 + 8^4 = 6479 ≈ 9^4$
$9^5 + 10^5 = 159049 ≈ 11^5$
As you can see, there is a pattern occurring in the form of:
$(2n-1)^n + (2n)^n ≈ (2n + 1)^n \{n ∈ ℤ+; n > 0\}$
It also seems the that if you rearrange the equation and take the limit at +∞:
$\lim_{n\to +\infty} { [(2n-1)^n + (2n)^n]^{1/n} - (2n + 1) } = 2ln(1 + e^{1/2}) - 2 ≈ -0.051846 $
(Calculated using Wolfram|Alpha)
Has this been noticed before; if so, is there a proof or is this the proof?

REPLY [8 votes]: $$\left(\dfrac{2n-1}{2n+1} \right)^n = \left(1 -\dfrac2{2n+1} \right)^n \sim \dfrac1e - \dfrac1{en^2} + \mathcal{O}(1/n^4)$$
$$\left(\dfrac{2n}{2n+1} \right)^n = \left(1 -\dfrac1{2n+1} \right)^n \sim \dfrac1{\sqrt{e}}$$
Hence,
$$\left(\dfrac{2n-1}{2n+1} \right)^n + \left(\dfrac{2n}{2n+1} \right)^n \sim \dfrac1e + \dfrac1{\sqrt{e}} \approx 0.97441$$