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

 A: As an alternative, we can directly check $f(n)< f(n+1)$ that is
$$\frac{n}{2n+1}<\frac{n+1}{2n+3} \iff 2n^2+3n<2n^2+3n+1 \iff0<1$$
which is true.
A: We avoid using induction in this approach.
You have $$f(n)=\frac{n}{2n+1}$$
Hence, $$f(n)=\frac{1-\frac{1}{2n+1}}{2}$$
Now, as $n$ increases, $\frac{1}{2n+1}$ decreases and hence,
$f(n)$ increases.
A: $\dfrac{n}{2n+1} = \dfrac{1}{2} \left( \dfrac{2n}{2n+1} \right) = \dfrac{1}{2} \left( \dfrac{2n+1 - 1}{2n+1} \right) = \dfrac{1}{2} \left(1 - \dfrac{1}{2n+1} \right)$
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 $n$.
A: A proof by induction would express that if $f$ is increasing at $n$, then $f$ is increasing at $n+1$, or
$$\frac n{2n+1}<\frac{n+1}{2n+3}\implies\frac{n+1}{2n+3}<\frac{n+2}{2n+5},$$
which can be rewritten as
$$-\frac1{(2n+1)(2n+3)}<0\implies-\frac1{(2n+3)(2n+5)}<0.$$
This proposition is true, but a little nonsense as both members are tautologies.
A: As Kavi Rama Murthy and "user" point out, you don't need induction to show that $f(n + 1) \geq f(n)$.  But (according to the usual definition of "increasing"), you need to show more than that; that for any $m$ and $n$ with $n \geq m$, $f(n) \geq f(m)$.  And to derive that rigorously from $f(n + 1) \geq f(n)$, you do need induction (even though the inference is intuitively obvious).
But you can show $n \geq m \ \Rightarrow f(n) \geq f(m)$ without induction.  GeometryLover and ilovemath deduce that from the premise that the sequence $n \mapsto \frac{1}{2n + 1}$ is decreasing.  More directly, we have
\begin{align*}
f(n) - f(m)
&= \frac{n}{2n + 1} - \frac{m}{2m + 1}\\
&= \frac{n - m}{(2m + 1)(2n +1)}\\
&> 0,
\end{align*}
so $f(n) > f(m)$, whenever $n > m$.
