Proving a function is increasing in $n$. I am trying to prove that the function $\frac{n}{2n+1}$, defined for $n \in \mathbb{N}$, decreases in $\mathbb{N}$. I attempted it by induction, but I'm not convinced that I fully need induction. Why can I not prove that for an arbitrary $n$, $f(n) \leq f(n+1)$ and deduce that, because $n$ was arbitrary, this holds for all $n$? The only thing left out would be the base case, but. I'm not fully sure why I need it here.
Regardless, here is my attempt at the induction:

Let $f: \mathbb{N} \to \mathbb{R}$ be defined by $f(n) = \frac{n}{2n+1}$. We prove by induction on $n$ that $f$ is increasing in $n$. If $n = 1$, we notice that
\begin{align*}
f(1) = \frac{1}{3} \leq \frac{2}{5} = f(2). 
\end{align*}
Suppose inductively that we have $f(n) \leq f(n+1)$ for some $n \geq 1$. So we have
$\frac{n}{2n+1} \leq \frac{n+1}{2n+3}$. First, we have
\begin{align*}
\frac{n+1}{2n+3} \leq \frac{n+3}{2n+3}.
\end{align*}
Furthermore, $2n + 5 \geq 2n + 3$, so $\frac{1}{2n + 5} \leq \frac{1}{2n+3}$, so $\frac{n+3}{2n + 3} \leq \frac{n+3}{2n+5}$. Therefore, it follows that
\begin{align*}
\frac{n+1}{2n+3} \leq \frac{n+3}{2n+3} \leq \frac{n+3}{2n + 5} = \frac{(n+2) + 1}{2(n+2) + 1}, 
\end{align*}
so $f(n+1) \leq f(n+2)$, which closes the induction


$\dfrac{n}{2n+1} = \dfrac{1}{2} \left( \dfrac{2n}{2n+1} \right) = \dfrac{1}{2} \left( \dfrac{2n+1 - 1}{2n+1} \right) = \dfrac{1}{2} \left(1 - \dfrac{1}{2n+1} \right)$
Since $ \dfrac{1}{2n+1} $ decreases with increasing $n$, then $ \left( 1 - \dfrac{1}{2n+1} \right)$ is increasing with $n$, thus $\dfrac{n}{2n+1} $ increases with increasing $n$.
