The limit of a sequence reveals its long-term behavior. In this article, we will provide a step-by-step guide to determining the limit of a sequence.
Step 1: Does the Sequence Converge?
The first step to finding the limit of a sequence is to figure out if it converges or not. For example, the sequence \(a_n = \sin n\) does not converge as \(n \to \infty\), because it oscillates and its terms do not approach any single value.
Another example is \(a_n = \dfrac{n^n}{n!}\), which diverges because its terms increase without bound. You can see this intuitively, since $n^n$ increases faster than $n!$, but if you want to prove it, you’ll have to use the Ratio Test:
\[ \begin{align*} \lim_{n\to\infty}\frac{a_{n+1}}{a_n} &= \lim_{n\to\infty}\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}\\ &=\lim_{n\to\infty}\frac{(n+1)n!}{(n+1)!}\cdot\frac{(n+1)^n}{n^n}\\ &= \lim_{n\to\infty}1\cdot\left(1 + \frac1n\right)^n = e > 0. \end{align*} \]
As we can see, figuring out whether a sequence converges can be a difficult task, but the good news is, if it diverges, we have our answer instantly. However, if it converges, we need to move on to step 2.
Step 2: What Does the Sequence Converge to?
This is the second and last step to figure out what a sequence converges to, and it is the more difficult of the two. This step is very difficult to give advice for, as it depends heavily on what the sequence is. You just have to use the limit rules — remember that we can take the limit of the terms of a sequence. For example, let’s say we have the sequence \(a_n = \dfrac{n^3 + 4n^2 – 7n + 6}{n^4-3n^2+15}\). We know, by the Dominant Term Rule, that this sequence will converge to \(0\), since the degree of the denominator is greater than the degree of the numberator. Of course, there are many different methods to evaluate a limit, the most powerful of which is L’Hopital’s Rule, which we will not go into here. You just have to use your ingenuity and the different limit rules as needed depending on the sequence at hand.
Examples
1) $\displaystyle \lim_{n\to\infty} a_n = \dfrac{n!}{e^n}$
We can see that this sequence will diverge to $+\infty$ because $\dfrac{a_{n+1}}{a_n} = \dfrac{n+1}{e}$, and for all $n > e$ this ratio is greater than $1$, meaning we can compare this sequence to a geometric sequence with ratio greater than $1$.
2) $\displaystyle \lim_{n\to\infty} b_n = \dfrac{n*2^n}{e^n}$
In this case, the sequence does converge, this time to $0$, as the ratio between the terms approaches $\dfrac{2}{e} < 1$ as $n$ goes to $\infty$.
Conclusion
These are the general 2 steps. You will need to make adjustments for individual practice problems. Make sure you understand the logic behind these steps.
