In the last exercise you saw that you can use the product rule for limits and the standard limit $\displaystyle\lim_{n\to\infty} \frac{1}{n} = 0$ to prove

$$\lim_{n\to\infty} \frac{1}{n^2} = \lim_{n\to\infty} \frac{1}{n} \cdot \lim_{n\to\infty} \frac{1}{n} = 0\text.$$ But in order to prove that $\displaystyle \lim_{n\to\infty} \frac{1}{\sqrt n} = 0$ we needed to assume the convergence of the sequence. Take a look at the proof below.

The following theorem states that the sequence $\frac{1}{\sqrt n}$ indeed converges.

In our case, the sequence $1, \tfrac{1}{\sqrt 2}, \tfrac{1}{\sqrt 3},\dots$ is descending (the terms are getting smaller), and the sequence is bounded by $1$.

For $k = 0$ we get the constant sequence $1, 1, 1, \dots$. The limit of this sequence is $1$ .

For $k = -1$ the sequence becomes

$$1,2,3,4,5,\dots$$ There is no real number $L$ such that $a_n \to L$. The sequences is greater than any finite number if you look far enough in the sequence. In our old definition, this sequence does not converge to any limit. But more can be said about this.

The $M$ in the definition can be seen as a large number that despite being so large is not an upper bound for the sequence. A large number is $M$ given, and then this definition gives an $N$ such that the sequence lies entirely above $M$ after that point. This definition captures the notion of a sequence that grows very large formally.

We denote this as

$$\lim_{n\to\infty} a_n = \infty\text.$$ Note that this does not mean that $|a_n - \infty|$ is small as in the original definition of a limit because you cannot perform calculations with $\infty$ as if it is an ordinary number.

The following calculation rule summarises everything:

We also consider this family of limits as standard limits. From here on you can use them now without proof.