Find a sequence limit
Dots that settle on a line
Each dot shows one term of the sequence \(a_n=\frac{3n+1}{n+2}\), and the dot at position \(n\) has the height \(a_n\).
Drag the position \(n\) to the right. The term rises from \(\frac43\) at \(n=1\) to about 2.58 at \(n=10\) and about 2.90 at \(n=50\).
The orange segment from the dot to the dashed line at height 3 shows the gap \(3-a_n=\frac{5}{n+2}\), which shrinks as \(n\) grows.
The orange strip around the line holds the heights within a chosen distance of 3, and the dots inside it are teal. Narrow the strip to half-width 0.1. Every dot from \(n=49\) on lies inside it, because the gap \(\frac{5}{n+2}\) is less than 0.1 when \(n+2>50\).
A narrower strip needs a larger \(n\), but a larger \(n\) exists for every strip.
Every strip catches the dots
The dashed line is at a height \(L\), and the orange strip holds the heights within a chosen distance of \(L\).
Narrow the strip. The faint dashed line marked \(N\) moves to the right, and the dots from \(N\) on still lie inside the strip.
If this happens for every width of the strip, the terms come as close to \(L\) as you choose, and the sequence converges to \(L\).
If the dots do not settle on one height, no strip holds all of them from some position on, so the sequence has no limit.
Notation
When the dots settle at the height \(L\), the sequence converges to \(L\), and \(L\) is its limit.
\[\lim_{n\to\infty}\cy{a_n}=\cy{L}\]If \(a_n=f(n)\) and \(f(x)\) has a limit as \(x\) grows, the sequence has the same limit, so you can use the limit methods for functions.
\[\lim_{n\to\infty}f(n)=\lim_{x\to\infty}f(x)\]L'Hôpital's rule needs a function of a real variable. Replace \(n\) by \(x\) before you differentiate.
Two limits come up often, and each holds under the condition beside it.
\[\begin{aligned}\lim_{n\to\infty}\dfrac{1}{n^p}&=0\ (p>0)\\ \lim_{n\to\infty}r^n&=0\ (|r|<1)\end{aligned}\]A sequence that does not converge diverges. It can diverge because the terms grow without bound, or because the terms do not settle on one number.
\[\begin{aligned}\lim_{n\to\infty}n^2&=\infty\\ (-1)^n&\ \text{has no limit}\end{aligned}\]
When it fails
Make the terms alternate in sign. The even terms approach 3 and the odd terms approach \(-3\), so the dots stay between \(-3\) and 3 but do not settle.
No strip around one height holds all the dots from some position on, so the sequence has no limit and diverges. Staying between two numbers is not enough for convergence.
Example 1: A ratio of polynomials
Find \(\displaystyle\lim_{n\to\infty}\frac{8n^2-3}{2n^2+5n}\).
Divide the top and the bottom by \(n^2\), the highest power of \(n\) in the fraction. The value of the fraction does not change.
\[\frac{8n^2-3}{2n^2+5n}=\frac{8-\frac{3}{n^2}}{2+\frac{5}{n}}\]As \(n\) grows, the terms \(\frac{3}{n^2}\) and \(\frac5n\) approach 0, so the top approaches 8 and the bottom approaches 2.
\[\frac{8-\frac{3}{n^2}}{2+\frac{5}{n}}\ \to\ \frac{8-0}{2+0}\]The terms approach \(\frac82=4\). The labeled dots rise from \(\frac57\) at \(n=1\) to about 3.81 at \(n=50\), toward the dashed line at height 4.
\[\lim_{n\to\infty}\frac{8n^2-3}{2n^2+5n}=4\]
Example 2: A power against an exponential
Find the limit of the sequence \(a_n=\dfrac{n}{e^n}\).
The first terms are about 0.37, 0.27, 0.15, 0.07 and 0.03. The dots fall toward the axis.
A tempting wrong move: Both \(n\) and \(e^n\) grow without bound. A number divided by itself is 1, so the limit is 1.
\[\frac{\infty}{\infty}=1\]A limit of 1 needs dots near the red dashed line at height 1, but no dot is above 0.4. The top and the bottom grow at different speeds, so \(\frac{\infty}{\infty}\) has no fixed value.
Replace \(n\) by \(x\). The function \(\frac{x}{e^x}\) has the form \(\frac{\infty}{\infty}\), so you can use L'Hôpital's rule and differentiate the top and the bottom.
\[\lim_{x\to\infty}\frac{x}{e^x}=\lim_{x\to\infty}\frac{1}{e^x}=0\]The limit of the function is 0, so the sequence converges to 0.
Practice
Decide whether the sequence \(a_n=\dfrac{3n^2-n}{n+5}\) converges. If it converges, find the limit.
Move the slider to large \(n\). Decide whether the dots level off at one height or rise without end.
Solution
Divide the top and the bottom by \(n\), the highest power of \(n\) in the bottom.
\[\frac{3n^2-n}{n+5}=\frac{3n-1}{1+\frac5n}\]The bottom approaches 1, but the top, \(3n-1\), grows without bound, so the terms grow without bound too.
\[\lim_{n\to\infty}a_n=\infty\]The terms have no finite limit, so the sequence diverges.
Questions
Questions 1 and 2 use the figure. It shows the dots for \(a_n=4+\frac{2(-1)^n}{n}\), for \(n=1\) to 12. The dashed line is at height 4. The orange strip covers the heights from 3.4 to 4.6, and the dots outside it have a red ring.
What is the limit of the sequence?
- 4
- 5
- The limit does not exist, because the dots go above and below the line
Answer
4
Which is the smallest \(n\) such that the dots for \(n\), \(n+1\), \(n+2\), and so on all lie inside the orange strip?
- 4
- 2
- 3
- 5
Answer
4
What is \(\displaystyle\lim_{n\to\infty}\left(\frac23\right)^n\)?
- 0
- \(\dfrac23\)
- 1
Answer
0
Summary
You find the limit of a sequence by finding the limit of its formula as the position grows.
The terms of a convergent sequence come as close to one number as you choose, and the limit of the formula gives that number.
You use it to decide whether a sequence converges, grows without bound or fails to settle.
This is one of five skills in topic 10.1 of Which & Why: AP Calculus.