Behavior of the sequences as n tends to infinity

Question

I'm trying to figure out the behavior of some sequences in the following cases.

The next two can be found in Hardy's "A Course of Pure Mathematics" (p. 131).

If $\phi(n) \to +\infty$ and $\psi(n) \to -\infty$, then

1) $\lim_\limits{n \to +\infty} (\phi(n) + \psi(n)) = 0$ if $\phi(n) = n$ and $\psi(n) = -n$

2) $\lim_\limits{n \to +\infty} (\phi(n) + \psi(n)) = \infty$ if $\phi(n) = n^2$ and $\psi(n) = -n$

In the first case, if $n-n$ were simply numbers, then I could substract them and obtain a constant sequence which has the same limit. However, I'm considering this case as a limit, so I'm not quite sure I can operate with the expression under the limit as with simple numbers.

In the second case, I could make the following transformation $\lim_\limits{n \to +\infty} (n^2-n) = \lim_\limits{n \to +\infty} n^2 \cdot (1-\frac{1}{n})$, but as far as I understand, I can't use the property $\lim_\limits{n \to +\infty} x(n) \cdot y(n) = \lim_\limits{n \to +\infty} x(n) \cdot \lim_\limits{n \to +\infty} y(n)$ because $\lim_\limits{n \to +\infty} n^2 = +\infty$. So it's like in the case of, for example, $\lim_\limits{n \to +\infty} (\sqrt{n^2+n} - n)$ where the similar transformation $\lim_\limits{n \to +\infty} (n \cdot \sqrt{1+\frac{1}{n}} - n)$ leads me to the same indeterminate form $"\infty-\infty"$ (but here I can get rid of it with help of the conjugate multiplier $(\sqrt{n^2+n} + n$)).

So, I'm confused of how can I deal with the indeterminate form of type $"\infty-\infty"$ and would like to find a way to solve this by only using the basic properties of the limits.

Thanks!

Answer 1

You may add the functions, so that $\phi(n) + \psi(n) = n - n = 0$ for all $n$. Then take the limit. That is exactly what $$\lim_{n\to\infty}( \phi(n)+\psi(n) ) $$ is defined to mean.

In the second case, again you are right:
$$\lim_{n\to\infty} (\phi(n) + \psi(n)) = n^2 - n. $$

At this stage you do not need to apply the limit of a product but simply note that once $n > 2$, $n^2 - n > n$ so the right hand side approaches $\infty$.

Does that answer your question?

Answer 2

I will provide an answer with aims of explaining why the approach suggested in the comments does not work every time. I present three observations:

1. Let $\lim_{n\to\infty}f(n)\to\infty$ and $\lim_{n\to\infty}g(n)=c>0$. Then:

$$\lim_{n\to\infty}f(n)g(n)\to c\lim_{n\to\infty}f(n) \to \infty$$

(I can provide a formal proof below, if needed).

2. Let $\lim_{n\to\infty}f(n)\to\infty$ and $\lim_{n\to\infty}g(n)=0$. Then we cannot be conclusive about $\lim_{n\to\infty}f(n)g(n)$, since it is of the form $0\cdot \infty$.

3. Consider the function

   $$r(n;k)=n\sqrt{c+1/n}-n=n(\sqrt{k+1/n}-1)=f(n)\cdot g(n)$$

Where $f(n)=n$ and $g(n)=\sqrt{c+1/n}-1$. Notice that $\lim_{n\to\infty}f(n)\to\infty$ and $\lim_{n\to\infty}g(n)=\sqrt{k}-1$.

This means that for $k>1$, it is the case that:

$$\lim_{n\to\infty}f(n)g(n)\to (\sqrt{k}-1)\lim_{n\to\infty}f(n) \to \infty$$.

However, if $k=1$ the limit is of the form $0\cdot\infty$ and hence:

$$\lim_{n\to\infty}f(n)g(n)\neq (\sqrt{k}-1)\lim_{n\to\infty}f(n) =0$$.

Notice that $k=1$ represents the counter example you are providing, however for $n^2-n=n^2(1-1/n^2)$ this problem does not arise since $\lim_{n\to \infty}(1-1/n^2)=1\neq 0$ and hence

$$\lim_{n\to \infty}n(1-1/n^2)\to 1\cdot \lim_{n\to \infty}n\to \infty$$