Prove identities for sums of series using the principle of mathematical induction

What this dot point is asking

NESA wants you to use mathematical induction to prove that a formula for a finite sum is correct for every positive integer $n$, not just for the few values you could check by hand. A series identity is a claim of the form "this running total equals this closed expression", and induction is the natural tool because the running total at $n = k + 1$ is built directly out of the running total at $n = k$. The proof always has the same four named parts, and the marks are spread across all four, so reproducing the structure exactly is half the task.

The answer

The principle of mathematical induction

The principle of mathematical induction lets you prove infinitely many statements at once from two finite checks. Let $P(n)$ be a statement about a positive integer $n$. If

1. $P(1)$ is true (the base case), and
2. for every positive integer $k$, the implication $P(k) \implies P(k + 1)$ holds (the inductive step),

then $P(n)$ is true for all integers $n \ge 1$.

The picture to keep in your head is a row of dominoes. The base case knocks the first one over; the inductive step guarantees that whichever domino falls knocks over the next; together they topple the whole infinite chain.

Two checks, not infinitely many: that is the whole power of the method. Notice that the base case may start at $n = 0$ or some other integer rather than $n = 1$ if that is what the statement asserts; start the chain at whatever the smallest claimed value of $n$ is.

The standard four-part structure

The four-part structure is the template every series proof must follow, and you should write each part with its own heading or label so the marker can tick them off.

Part 1: Base case

Verify $P(1)$ by substituting $n = 1$ into both sides separately and checking they are equal. Compute the two sides independently; never assume they match.

Part 2: Inductive hypothesis

Assume $P(k)$ holds for some positive integer $k$, and write the assumed identity out in full. This line is the fact you are allowed to use later, so make it explicit.

Part 3: Inductive step

Prove $P(k + 1)$ using the hypothesis. For a series, the move is mechanical: write the $(k + 1)$-term sum as the $k$-term sum plus the new last term, replace the $k$-term sum by the formula the hypothesis gives, then simplify to the closed form with $n = k + 1$ substituted.

Part 4: Conclusion

State that since the base case holds and $P(k) \implies P(k + 1)$, by the principle of mathematical induction $P(n)$ holds for all positive integers $n$.

Common series formulas worth memorising

The four standard sum formulas come up constantly, both as the thing you must prove and as a result you can quote elsewhere:

You should be able to prove each of these on demand. A neat sanity check: the sum of cubes equals the square of the sum of integers, so $\sum i^3 = \left( \sum i \right)^2$.

The split-off-the-last-term technique

The technique that drives the inductive step is always the same: peel off the final term so the rest matches the hypothesis. In symbols,

So the recipe for Part 3 is:

1. Write the LHS for $n = k + 1$ and split off the $(k + 1)$-th term.
2. Replace the remaining $k$-term sum by the closed form from the hypothesis.
3. Factor out the common factor (almost always $(k + 1)$) and simplify until you reach the target formula with $n = k + 1$ in place of $n$.

If the algebra gets messy, write down the target RHS at $n = k + 1$ first and steer your working towards it; knowing where you are heading makes the factoring obvious.

How exam questions ask about series identities

• "Prove by mathematical induction that $\sum \dots = \dots$": the direct instruction. Reproduce all four parts; the verb "induction" tells you no other method earns the marks.
• "Prove that $S_n = \dots$ for all positive integers $n$": the same task even though "induction" is not named, when $S_n$ is a sum. Use the four-part structure.
• "Hence find $\sum_{i = 1}^{50} \dots$": a follow-up after you have proved the identity; substitute $n = 50$ into the closed form you just proved.
• "Show that the result holds for $n = 1, 2, 3$, then prove it in general": the first part is the base case (plus two extra checks for confidence); the "in general" part is the full induction.
• "Prove $\sum_{i = 1}^{n} \frac{1}{i(i + 1)} = \frac{n}{n + 1}$": a telescoping or partial-fraction sum. Induction works exactly as for polynomial sums; split off the last term and simplify.

Edge cases and what-ifs

• A base case that is not $n = 1$. If the identity is only claimed for $n \ge 2$ (for instance because a term is undefined at $n = 1$), verify the base case at $n = 2$ and run the step from $k \ge 2$. The logic is unchanged.
• Checking your closed form first. If you are unsure of the formula, evaluate the sum at $n = 1, 2, 3$ and confirm the closed form matches before you commit to a proof. The base case only checks one value, so a wrong formula can still pass it.
• Telescoping as a cross-check. A sum like $\sum \frac{1}{i(i + 1)}$ also collapses directly because $\frac{1}{i(i + 1)} = \frac{1}{i} - \frac{1}{i + 1}$. Induction is still the method the question asks for, but the telescoped value is a fast way to confirm your target RHS is right.
• The empty sum. A sum from $i = 1$ to $i = 0$ is empty and equals $0$. This is why a base case at $n = 0$ for $\sum_{i = 1}^{n} i = \frac{n(n + 1)}{2}$ also works: both sides are $0$.