2
$\begingroup$

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 $a_n=f(n)0}$. 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.

3 Answers 3

3

It's perfectly fine to use $f(x)$ to conclude that the sequence $\{a_n\}$ is increasing, with one caveat: if your class hasn't yet introduced differentiation and proved that a function with a positive derivative is increasing, then the professor may object to you using this fact without proof.

In your particular example, I think the easiest solution is to observe that $a_n=\frac{n}{n+1}=1-\frac{1}{n+1}$. Proving that $\{a_n\}$ is increasing is therefore equivalent to proving that the sequence $\{\frac{1}{n+1}\}$ is decreasing, which is much easier to show.

  • 1
    I didn't mention this, but since first year calculus is a pre-requisite for this course (in which we showed the relationship between increasing functions and their derivative) I presume we can use this fact. Thank you for the answer, and that method of solving the problem by showing $(\frac{1}{n+1})$ is decreasing (much easier!). – 2017-01-18
1

It all depends on what is considered "known". When you have studied sequences, but not yet differentiation of real functions, the second method would not be accepted. But after you have taken calculus, the second method is OK.

  • 0
    ... In particular courses and textbooks (at least the good ones) build on known results starting from axioms, and by the time one encounters such exercises differentiation is almost never studied. But in higher mathematics it is customary to study sequences using calculus and complex analysis. – 2017-01-18
0

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.