Using an Increasing Differentiable Function to Show A Sequence is Increasing I am taking my first real analysis course, and we are talking about sequences of real numbers. I have a question where as part of a proof I want to show that a sequence $a_n=\frac{n}{n+1}$ for $n\in\Bbb Z_{>0}$, is strictly increasing. Note $\Bbb Z_{>0}=\{1,2,3,\ldots\}$. I understand that induction is often used for showing these types of things, since $n$ is a positive integer. Another method that I used was showing $\frac{a_{n+1}}{a_n}> 1$: since $\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{n(n+2)}$, we have
$$(n+1)^2=n^2+2n+1>n^2+2n=n(n+2)\implies \frac{(n+1)^2}{n(n+2)}>1$$
However, as I am very familiar with real valued functions of real variables, and derivatives, I thought I could show $(a_n)$ is strictly increasing by showing that the function $f:\Bbb R_{>0}\to\Bbb R$ defined as $f(x)=\frac{x}{x+1}$ is strictly increasing. Note $\Bbb R_{>0}=\{x\in\Bbb R:x>0\}$. Surely $f$ is differentiable on $(0,\infty)$, so $f'(x)=\frac{1}{(x+1)^2}>0$ for all $x>0$. Thus $f$ is strictly increasing for $x>0$. Since $f$ is strictly increasing for $x>0$, we can conclude that $a_n=f(n)<f(n+1)=a_{n+1}$ for every $n\in\Bbb Z_{>0}$. Thus, $(a_n)$ is strictly increasing.
$\mathbf{My\,Question:}$ is the method of solving this problem that I showed in the previous paragraph (i.e. using $f(x)$) a valid and accepted method for showing $(a_n)$ is increasing. The question is really about any $(a_n)$ where we can construct a differentiable function such that $f(n)=a_n$, this is just one example. I question whether it is valid, because we are taught to use induction and the method of showing $\frac{a_{n+1}}{a_n}> 1$ to show that sequences are strictly increasing. Although I find the method of using $f(x)$ to be much easier in some cases (and quite intuitive), yet I never see it in lectures.
Thanks in advance for any help. 

Suppose we manage to construct, $f$ such that $f(n)=a_n$ and $ \forall x >0, f'(x) >0$, then we know that $f$ is an increasing function.
That is $x>y$ then $f(x)> f(y)$.
In particular since $n+1 > n$, we have shown that $a_{n+1} > a_n$ and hence the sequence is increaseing.
(remark: if we have further information such as $a_n>0$, we have just proven that $\frac{a_{n+1}}{a_n}>1$)
But I agree with other answers that it depends on what is known to you. 
