There is a test problem which asks you to define a non-recursive formula for this sequence: $$\{\frac21, \frac44, \frac69, \frac8{16}, \frac{10}{25}, \ldots \}$$
I saw the relationship $a_n = \frac{2n}{n^2}$, which I simplified to $\frac2n$ and marked that as my final answer. This was marked wrong as while $\frac2n$ gave the exact same values, it produced a sequence that is $\{\frac21, \frac22, \frac23, \frac24, \frac25, \ldots\}$ instead of the above sequence. Even though the numbers are the same, my teacher argues that the simplification makes the sequences different.
She asks for a mathematical proof or a logical statement that these two sequences are the same. She says she is open to reconsidering if she reads a logical and mathematical argument by a mathematician.
Is $\frac2n$ right? If it is, can you provide a mathematical proof for that?
$\require{cancel}$ You have proven it.
$$\frac{2n}{n^2}=\frac{2\cdot {n}}{n \cdot {n}}=\frac{2\cdot \cancel{n}}{n \cdot \cancel{n}}=\frac{2}{n}.$$
The cancelation is valid since $n>0$.
The sequences $$\{\frac21,\frac44,\frac69,\frac8{16},\frac{10}{25},\ldots\}=\{2,1,\frac23,\frac12,\frac25,\ldots \}=\{\frac21,\frac22,\frac23,\frac24,\frac25,\ldots\}$$
because they are termwise equal.
Credit: Hagen vonEitzen's comment.