Nth Terms and Limits of Sequences Revision Notes | GCSE AQA Higher Further Maths

1. What nth Terms Are and Why They Matter

Imagine you need the 200th term of a sequence. Writing out 200 terms is not practical. An nth term formula lets you jump straight to any term by substituting the position number $n$.

The nth term is an algebraic rule: substitute $n = 1$ to get the first term, $n = 2$ for the second, and so on. For example, the nth term $3n + 1$ gives the sequence $4, 7, 10, 13, \ldots$

There are two core skills here: using a given nth term to find specific terms, and finding the nth term from a sequence. Both appear regularly in GCSE Further Maths papers.

2. Using the nth Term to Find Specific Terms

When you are given the nth term formula, finding any term is a substitution exercise. The key discipline is to substitute carefully and simplify one operation at a time.

Worked example: Using the nth Term: Finding Terms from a Formula

The nth term of a sequence is $u_n = \dfrac{4n - 1}{n + 2}$. Find the 1st, 3rd, and 10th terms.

1. $u_1 = \dfrac{4(1) - 1}{1 + 2} = \dfrac{3}{3} = 1$ — Replace every $n$ with 1, then evaluate numerator and denominator separately. [M1]
2. $u_3 = \dfrac{4(3) - 1}{3 + 2} = \dfrac{11}{5} = 2.2$ — Numerator: $4 \times 3 - 1 = 11$. Denominator: $3 + 2 = 5$. [A1]
3. $u_{10} = \dfrac{4(10) - 1}{10 + 2} = \dfrac{39}{12} = \dfrac{13}{4} = 3.25$ — Numerator: $40 - 1 = 39$. Denominator: $10 + 2 = 12$. Simplify the fraction. [A1]

Answer: $u_1 = 1$, $\quad u_3 = \dfrac{11}{5}$, $\quad u_{10} = \dfrac{13}{4}$

Notice how the terms $1, 2.2, 3.25, \ldots$ are increasing but appear to be getting closer together. That pattern is the first hint that this sequence has a limiting value — something we will return to shortly.

3. Finding the nth Term of a Sequence

Going the other way — from a list of terms to a formula — is equally important. For a linear (arithmetic) sequence, the method is systematic: find the common difference, then adjust the constant.

Worked example: Finding the nth Term of a Linear Sequence

Find the nth term of the sequence $7, 11, 15, 19, \ldots$

1. $11 - 7 = 4, \quad 15 - 11 = 4, \quad 19 - 15 = 4$ — The difference between consecutive terms is constant at 4, so the nth term starts with $4n$. [M1]
2. $4(1), 4(2), 4(3), 4(4) = 4, 8, 12, 16$ — Compare this with the original sequence $7, 11, 15, 19$.
3. $7 - 4 = 3, \quad 11 - 8 = 3, \quad 15 - 12 = 3$ — Each term in the original sequence is 3 more than the corresponding $4n$ term. [M1]
4. $u_n = 4n + 3$ — Check: $u_1 = 4(1) + 3 = 7$ ✓, $u_4 = 4(4) + 3 = 19$ ✓. [A1]

Answer: $u_n = 4n + 3$

The same logic applies to sequences with a negative common difference. If the terms decrease by 3 each time, the nth term starts with $-3n$, and you adjust the constant in exactly the same way.

4. The Limiting Value of a Sequence

Some sequences do not grow forever — their terms home in on a fixed value. That fixed value is called the limiting value, and finding it is the second major skill in this topic.

The technique for finding a limit relies on one key observation: as $n \to \infty$, any term with $n$ in the denominator tends to zero. So the strategy is to divide numerator and denominator by the highest power of $n$ present, then let those terms vanish.

Worked example: Finding the Limiting Value of a Sequence

Find the limiting value of the sequence with nth term $u_n = \dfrac{4n - 1}{n + 2}$ as $n \to \infty$.

1. Highest power is $n^1$ (linear in both numerator and denominator). — This tells us to divide everything by $n$. [M1]
2. $u_n = \dfrac{\dfrac{4n}{n} - \dfrac{1}{n}}{\dfrac{n}{n} + \dfrac{2}{n}} = \dfrac{4 - \dfrac{1}{n}}{1 + \dfrac{2}{n}}$ — Every term is now expressed with $\frac{1}{n}$ factors, which will tend to zero. [M1]
3. As $n \to \infty$, $\dfrac{1}{n} \to 0$ and $\dfrac{2}{n} \to 0$ — Any constant divided by an ever-growing $n$ shrinks towards zero.
4. $\lim_{n \to \infty} u_n = \dfrac{4 - 0}{1 + 0} = 4$ — The sequence converges to 4. [A1]
5. $u_{1000} = \dfrac{4(1000) - 1}{1000 + 2} = \dfrac{3999}{1002} \approx 3.991$ — Very close to 4, confirming the limit. This also matches the terms we found earlier getting closer together.

Answer: The limiting value is $4$.

This connects directly to the terms we calculated in Section 2: $u_1 = 1$, $u_3 = 2.2$, $u_{10} = 3.25$. Each term is creeping towards 4 without ever reaching it — that is exactly what convergence looks like.

5. Limits with Quadratic and Higher-Degree Terms

The divide-by-the-highest-power strategy works for any rational nth term, not just linear ones. When the numerator and denominator have different degrees, the result changes significantly.

Three cases are worth knowing: if the degree of the numerator is less than the degree of the denominator, the limit is 0; if the degrees are equal, the limit is the ratio of leading coefficients; if the numerator has a higher degree than the denominator, the sequence diverges (no finite limit).

Worked example: Limit When Degrees Differ

Find $\lim_{n \to \infty} u_n$ where $u_n = \dfrac{3n^2 + 5}{2n^2 - n}$.

1. Both numerator and denominator have highest power $n^2$. — Divide everything by $n^2$. [M1]
2. $u_n = \dfrac{\dfrac{3n^2}{n^2} + \dfrac{5}{n^2}}{\dfrac{2n^2}{n^2} - \dfrac{n}{n^2}} = \dfrac{3 + \dfrac{5}{n^2}}{2 - \dfrac{1}{n}}$ — Each term now has $\frac{1}{n}$ or $\frac{1}{n^2}$ factors.
3. $\dfrac{5}{n^2} \to 0$ and $\dfrac{1}{n} \to 0$ — Both extra terms vanish as $n$ grows without bound.
4. $\lim_{n \to \infty} u_n = \dfrac{3 + 0}{2 - 0} = \dfrac{3}{2}$ — Equal degrees: limit equals ratio of leading coefficients $\frac{3}{2}$. [A1]

Answer: $\lim_{n \to \infty} u_n = \dfrac{3}{2}$

6. Practice Problems

The best way to consolidate both skills — using nth terms and finding limits — is to work through varied examples. Try each problem before checking the solution.

7. Exam Technique and Common Mistakes

Both specification skills — using nth terms and finding limits — carry method marks as well as accuracy marks. Showing the divide-by-$n$ step explicitly is what earns the method mark, even if you make an arithmetic slip afterwards.

A common error is to substitute a large number (such as $n = 1000$) and call that the limit. This approach can give a good approximation, but it is not an algebraic proof and will not earn full marks. Always use the divide-by-highest-power method.