Series

A sequence is a list, $a_1, a_2, a_3, \dots$; a series is what you get when you replace the commas with plus signs:

where: each $a_k$ is a term of the underlying sequence. If the list stops after $n$ terms the series is finite; if it continues forever it is infinite. The distinction matters because a series and its value are two different things — the series is the indicated addition, while its value, once you actually add up the first $n$ terms, is called the partial sum.

Partial sums carry all the information. The sum of the first $n$ terms is written $S_n$:

where: $S_n$ is a single number, the $n$-th partial sum. Because $S_n$ already contains $S_{n-1}$ inside it — everything except the last term — subtracting recovers that last term exactly:

where: the $n \ge 2$ restriction exists because $S_0$, the sum of zero terms, is not a defined partial sum here, so $a_1$ must be read off separately as $S_1$ itself.

Sigma notation compresses the addition. Instead of writing out every term, the capital Greek letter $\Sigma$ ("sum") does the work:

where: $k$ is the index of summation, starting at the lower limit $1$ and running up to the upper limit $n$; $a_k$ is the rule for the $k$-th term. Reading a sigma expression is a two-step habit — first identify the rule $a_k$, then plug in every integer $k$ from the lower to the upper limit and add. Three standard sums built from this notation turn up throughout the chapter as quick-reference facts rather than results you are asked to re-derive on demand:

where: $n$ is the number of terms summed and $k$ runs from $1$ to $n$ in each case. Once a sequence turns out to be a specific type — an AP or a GP — these general series ideas specialise into the familiar sum formulae $S_n = \tfrac{n}{2}[2a+(n-1)d]$ for an AP and $S_n = \tfrac{a(r^n-1)}{r-1}$ for a GP, which the rest of the chapter is devoted to.

Problem. Write the series $\displaystyle\sum_{k=1}^{5} (2k+1)$ in expanded form and find its value; then, for a second sequence whose partial sums are known to be $S_n = 3n^2 + 2n$, find its first four terms.

Step 1 — expand the sigma notation. Substitute $k = 1, 2, 3, 4, 5$ into the rule $2k+1$: this gives $3, 5, 7, 9, 11$, so the series is $3+5+7+9+11$.

Step 2 — add the terms. $3+5=8$, then $8+7=15$, then $15+9=24$, then $24+11=35$, so

Step 3 — find $a_1$ for the second sequence directly, since $a_1 = S_1$: $a_1 = S_1 = 3(1)^2+2(1) = 5$.

Step 4 — use $a_n = S_n - S_{n-1}$ for the rest. $S_2 = 3(4)+2(2) = 16$, so $a_2 = 16-5 = 11$. $S_3 = 3(9)+2(3) = 33$, so $a_3 = 33-16 = 17$. $S_4 = 3(16)+2(4) = 56$, so $a_4 = 56-33 = 23$.

Final answer. $\displaystyle\sum_{k=1}^{5}(2k+1) = \mathbf{35}$, and the second sequence's first four terms are $\mathbf{5, 11, 17, 23}$ — note the constant common difference of $6$, a preview of how an AP's terms behave.

• Confusing a series with its sum. The series is the indicated addition $a_1+a_2+\dots$; $S_n$ is its numerical value. Mixing the two words up causes muddled proofs.
• Wrong limits in sigma notation. Starting the index at $k=0$ instead of $k=1$, or stopping one term short or long, silently changes the answer.
• Using $a_n = S_n - S_{n-1}$ at $n=1$. This only works for $n \ge 2$, since $S_0$ is not defined; $a_1$ must always be found separately as $a_1 = S_1$.
• Treating $\sum k$, $\sum k^2$, $\sum k^3$ as exam-derivation targets. At this level they are background facts to quote and use, not results you are expected to prove from scratch, unless a question explicitly asks for the proof.

Ask the tutor: if you want more practice converting between expanded sums and sigma notation, or extracting terms from a given $S_n$ formula, ask the AI tutor for extra worked problems.