# How do I find the limit of a sequence?

Sep 10, 2017

There is no general way of determining the limit of a sequence. Also, not all sequences have limits. However, if a sequence has a limit point, it must be unique. (This is an elementary result of analysis).

If a sequence be such that, the higher and higher terms get smaller in magnitude or alternatively the difference between consecutive terms decreases as the order of terms increase, you can have a limit. A pretty good example would be,

${\left\{\frac{1}{n}\right\}}_{n} = \left\{1 , \frac{1}{2} , \frac{1}{3} , \ldots .\right\}$

In this case the limit turns out to be $0$ because the higher the terms are, the closer they get to $0$ and for a sufficiently large $n$ which is infinity, $L i m \frac{1}{n} = 0$ which is the limit point.

Two things may be observed. First the limit point has to be unique for every sequence. Second, the limit point may not be a member of the sequence itself as in this case, $0$ does not represent any term of the ${\left\{\frac{1}{n}\right\}}_{n}$ sequence.

There can be other types of sequences such as the ones in which the consecutive terms increase in magnitude for higher values of $n$. In this case, difference between two consecutive terms increases and the sequence diverges altogether. A good example is,

${\left\{n\right\}}_{n} = \left\{1 , 2 , 3 , \ldots .\right\}$

Here, $L i m n = \propto$

There can be another type of sequence known as the oscillating sequence as shown,

${\left\{{\left(- 1\right)}^{n}\right\}}_{n} = \left\{- 1 , 1 , - 1 , 1 , \ldots .\right\}$

The series neither converges not diverges and is termed as oscillatory.