how does sqrt(n)/n equal 1/sqrt(n)?

I'm working through a probability book where the following is stated (chapter 3):

Definition 3.3

Let $a_n$ and $b_n$ be two sequences of numbers.

We say $a_n$ is asymptotically equal to $b_n$, and write $a_n \sim b_n$, if ...

$\lim_{n \to \infty} \frac{a_n}{b_n} = 1$

Example 3.4

If $a_n = n + \sqrt{n}$ and $b_n = n$ then, since $\frac{a_n}{b_n} = 1 + 1/\sqrt{n}$ and this ratio tends to 1 as $n$ tends to infinity, we have $a_n \sim b_n$

My math is very rusty (15+ years), so I'm confused how the following is arrived at:

$\frac{a_n}{b_n} = 1 + \frac{1}{\sqrt{n}}$

I expected it to be the following ...

$\frac{a_n}{b_n} = \frac{\sqrt{n} + n}{n} = 1 + \frac{\sqrt{n}}{n}$

Simple answers will be preferred over complex answers.

Edit:

Checking with sympy, it seems that both the book and my version are the same:

>>> a, b, n = symbols('a b n')
>>> a = n + sqrt(n)
>>> b = n    
>>> simplify( (a/b) - (1 + 1/sqrt(n)) ) == 0
True
>>> simplify( (a/b) - (1 + sqrt(n)/n) ) == 0
True

# Double check equality
>>> simplify( (1 + sqrt(n)/n) - (1 + 1/sqrt(n)) ) == 0
True

# and ...
>>> simplify( (sqrt(n)/n) - (1/sqrt(n)) ) == 0
True

# and ...
>>> simplify( (sqrt(n)/n) )
1 / sqrt(n)

# Double check I'm not doing something stupid
>>> simplify( (a/b) - (1) ) == 0
False

So my question becomes how can I derive my equation from the book equation.

Solution 1:

$n = \sqrt{n}^2$

So in your result you can simplify both numerator and denominator by $\sqrt{n}$ and get the book's result. Or, starting from the book's result, you just have to multiply both by $\sqrt{n}$.