Mastering Mathematics Smartly
by Wee Wen Shih

A unique self-help website that provides comprehensive coverage of mathematics at A-level & beyond, written in a student-friendly style.

Sequences & series

What is the relationship between a sequence and a series? What is the meaning of a convergent sequence/series or of a divergent sequence/series? If we have a convergent series, do we have a convergent sequence and vice versa?

Find out the answers from this summary.

Overcoming areas of difficulty

We will look at specific instances where students encounter difficulty and offer practical approaches to overcome them with ease.

Scenario 1: Sum to infinity.

Suppose we already know that \sum_{r = 1}^{n} \frac{1}{(2r + 1)(2r + 3)} = \frac{n}{3(2n + 3)}. How do we deduce the sum to infinity?

First, we note that the RHS expression is an improper fraction which can be simplified to \frac{1}{3\left(2 + \frac{3}{n}\right)}, by dividing numerator and denominator by n (a very useful strategy to remember and apply). Now as n \rightarrow \infty we see that \frac{3}{n} \rightarrow 0 and so the sum to infinity is \frac{1}{6}.

Scenario 2: Inequality.

Another way of looking at any sum is by means of an inequality. The improper expression can be written as \frac{1}{6} - \frac{1}{2(2n + 3)} by long division, and its value is always going to be less than \frac{1}{6} for any positive integer n. Thus the inequality \sum_{r = 1}^{n} \frac{1}{(2r + 1)(2r + 3)} < \frac{1}{6} holds.

Scenario 3: Relationships between sums (part 1).

Sometimes, students may be asked to deduce the sum to infinity of a new sum. Take \sum_{r = 1}^{\infty} \frac{1}{(2r + 3)^2} < \frac{1}{6} for example. The trick is to relate it to the sum to infinity of the old one.

First, we note that, for any positive integer r, (2r + 1)(2r + 3) < (2r + 3)^2.

Taking reciprocals, we have \frac{1}{(2r + 1)(2r + 3)} > \frac{1}{(2r + 3)^2}.

It then makes mathematical sense (why?) for us to conclude that \sum_{r = 1}^{\infty} \frac{1}{(2r + 1)(2r + 3)} > \sum_{r = 1}^{\infty} \frac{1}{(2r + 3)^2}, and we are done.

Scenario 4: Relationships between sums (part 2).

Suppose we have found the formula of \sum_{r = 1}^{n} \frac{(r - 1)^2}{2^{r + 1}} in one part of a question. How does one go on to deduce the formula of \sum_{r = 1}^{n} \frac{r^2}{2^r}?

An effective approach is to expand each sum and find a clear relationship between them.

Now \sum_{r = 1}^{n} \frac{(r - 1)^2}{2^{r + 1}} = \frac{0^2}{2^2} + \frac{1^2}{2^3} + \frac{2^2}{2^4} + ... + \frac{(n - 1)^2}{2^{n + 1}}

and \sum_{r = 1}^{n} \frac{r^2}{2^r} = \frac{1^2}{2^1} + \frac{2^2}{2^2} + \frac{3^2}{2^3} + ... + \frac{n^2}{2^n} .

By observation (can you see it?), we can deduce that \sum_{r = 1}^{n} \frac{r^2}{2^r} = 2^2 \sum_{r = 1}^{n + 1} \frac{(r - 1)^2}{2^{r + 1}}. We then substitute n with n + 1 in the formula of \sum_{r = 1}^{n} \frac{(r - 1)^2}{2^{r + 1}} to complete the solution.

Do note that this type of scenario may also appear in questions involving mathematical induction.