Seems Like They are Mistaken Twins!

First, let's rationalise $\frac{1}{n+\sqrt{n²+1}}$ .

$\frac{1}{n + \sqrt{n²+1}} \cdot \frac{n - \sqrt{n²+1}}{n - \sqrt{n²+1}}$

$= \frac{n - \sqrt{n²+1}}{n² - ( n² + 1)} = \frac{-n +\sqrt{n²+1}}{1}$ .

Then, we can simplify the equation

$n + \sqrt{n²+1} = 2n + (-n + \sqrt{n²+1} )$ .

$n + \sqrt{n²+1} = n + \sqrt{n²+1}$ .

Hence, it is $\boxed{\text{True}}$

Let x = n + sqrt(n^2 + 1). x>0. Is x = 2n + 1/x, or is x^2 - 2nx - 1 =0, 2x = 2n +/- sqrt(4n^2 + 4), x = n + sqrt(n^2 + 1), so true. Ed Gray

$n + \sqrt { n^2 + 1 } = 2n + \frac {1}{n + \sqrt { n^2 + 1 }}$

$\sqrt { n^2 + 1 } - n = \frac {1}{ \sqrt { n^2 + 1 } + n}$

$( \sqrt { n^2 + 1 } - n)( \sqrt { n^2 + 1 } + n ) = 1$

$n^2 + 1 - n^2 = 1$

$1 \equiv 1$

This means, that our equation is an identity (always true), whenever we can evaluate both sides.

We will show now, that for all real n (not just for positive real numbers) this identity stands. For this, it is enough to show that all real numbers are part of the domain of the RHS.

No negative number under the square root:

$\text {Since } \forall n \in \mathbb {R} : n^2 + 1 ≥ 1 ≥ 0 \Longrightarrow \sqrt { n^2 + 1 } \in \mathbb {R}$

No division by zero:

$\text {And since } \forall n \in \mathbb {R} : \sqrt { n^2 + 1} > \sqrt { n^2 } = | n | \Longrightarrow \sqrt { n^2 + 1 } + n > 0$

$\Longrightarrow \frac {1}{ n+ \sqrt { n^2 + 1} } \in \mathbb {R}$

$\text {Hence, our statement is } \boxed { \text {True} } \text { for all real values of n. }$