In short

Sigma notation \displaystyle\sum_{k=1}^{n} a_k is shorthand for a_1 + a_2 + \dots + a_n. Pi notation \displaystyle\prod_{k=1}^{n} a_k is shorthand for a_1 \cdot a_2 \cdots a_n. Both are controlled by an index variable, a starting value, and an ending value. Sigma notation obeys linearity rules (constants factor out, sums split), and telescoping sums — where consecutive terms cancel — are a powerful technique for evaluating sums in closed form.

Write out the sum of the squares of the first 50 positive integers. In longhand, that is 1^2 + 2^2 + 3^2 + 4^2 + \dots + 50^2 — a string of fifty terms connected by plus signs. Now imagine writing the sum of the first 500 terms, or a sum where the general term is a complicated expression like \frac{k^2}{(k+1)(k+2)}. Longhand becomes unreadable. You need a notation that says "add up these terms" without listing every one.

That notation is \Sigma (the Greek capital letter sigma, for "sum"). Instead of fifty terms separated by plus signs, you write:

\sum_{k=1}^{50} k^2

Five symbols replace fifty terms. The notation tells you three things: the index variable (k), where it starts (k = 1), and where it ends (k = 50). The expression after the sigma (k^2) is the general term — the formula that produces each term when you plug in successive values of k.

There is an analogous notation for products: \Pi (the Greek capital letter pi, for "product"). Where sigma adds, pi multiplies. The factorial n! = 1 \times 2 \times 3 \times \dots \times n is simply:

n! = \prod_{k=1}^{n} k

This article explains how to read, write, and manipulate both notations, and introduces telescoping sums — a technique that collapses certain sums into just two terms.

Anatomy of sigma notationA large sigma symbol in the centre with labels pointing to its parts. Below the sigma: k equals 1, labelled lower limit and index variable. Above the sigma: n, labelled upper limit. To the right: k squared, labelled general term. An annotation says this means add k squared for k from 1 to n. Σ k=1 n k² lower limit (start) upper limit (end) general term index variable = 1² + 2² + 3² + ... + n²
The four parts of sigma notation: the index variable $k$ names the counter, the lower limit sets where it starts, the upper limit sets where it stops, and the general term $k^2$ is the expression being summed. Replace $\Sigma$ with $\Pi$ and "sum" with "product" for pi notation.

Summation notation — the rules

Reading sigma notation

To evaluate \displaystyle\sum_{k=1}^{4} (2k + 1):

The index variable is a dummy variable — it does not matter whether you call it k, i, j, or m. The expressions \sum_{k=1}^{n} k^2 and \sum_{j=1}^{n} j^2 mean exactly the same thing.

Writing sigma notation

To express a sum in sigma notation, identify the pattern in the terms and write the general term as a formula involving the index.

The sum \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{99 \cdot 100} has general term \frac{1}{k(k+1)}, with k running from 1 to 99:

\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{99 \cdot 100} = \sum_{k=1}^{99} \frac{1}{k(k+1)}
Expanding sigma notation into individual termsAn arrow from the compact sigma notation on the left to the expanded form on the right. The sigma expression is the sum from k equals 1 to 5 of k squared. The expanded form shows 1 plus 4 plus 9 plus 16 plus 25 equals 55. Σ k=1 5 k² 1² + 2² + 3² + 4² + 5² = 1 + 4 + 9 + 16 + 25 = 55 compact form → expanded form → value
Sigma notation compresses a multi-term sum into a single expression. To evaluate it, expand by plugging in each value of the index, then add.

Properties of sigma notation

Sigma notation obeys several rules that make manipulation easier. These are not arbitrary conventions — they follow directly from the properties of addition.

Properties of Summation

For sequences a_k and b_k, a constant c, and integers 1 \leq m \leq n:

1. Constant multiple rule:

\sum_{k=1}^{n} c \cdot a_k = c \cdot \sum_{k=1}^{n} a_k

2. Sum/difference rule:

\sum_{k=1}^{n} (a_k \pm b_k) = \sum_{k=1}^{n} a_k \pm \sum_{k=1}^{n} b_k

3. Sum of a constant:

\sum_{k=1}^{n} c = nc

4. Splitting a range:

\sum_{k=1}^{n} a_k = \sum_{k=1}^{m} a_k + \sum_{k=m+1}^{n} a_k

These properties are the reason sigma notation is powerful: you can break a complicated sum into simpler pieces, factor out constants, and recombine. The rules are analogous to how you handle integrals — \int [f + g] = \int f + \int g — and this is no coincidence, since integration is the continuous analogue of summation.

Example of the rules in action. Evaluate \sum_{k=1}^{n} (3k^2 + 2k - 5):

\sum_{k=1}^{n} (3k^2 + 2k - 5) = 3\sum_{k=1}^{n} k^2 + 2\sum_{k=1}^{n} k - \sum_{k=1}^{n} 5
= 3 \cdot \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2} - 5n
= \frac{n(n+1)(2n+1)}{2} + n(n+1) - 5n

Each piece is a known formula. The sigma properties turned one complicated sum into three standard ones.

Splitting a sigma sum into partsA large box containing the sum of 3k squared plus 2k minus 5, with arrows splitting it into three smaller boxes: 3 times the sum of k squared, plus 2 times the sum of k, minus 5n. Each smaller box has its closed-form formula below. Σ(3k² + 2k − 5) 3·Σk² = n(n+1)(2n+1)/2 2·Σk = n(n+1) −5n = −5n + − Linearity: split, factor constants, use known formulas.
The sum $\sum(3k^2 + 2k - 5)$ splits into three standard sums by the linearity properties. Each piece has a known closed form, so the original sum can be evaluated without expanding every term.

Product notation

Pi notation does for multiplication what sigma does for addition. The expression \displaystyle\prod_{k=1}^{n} a_k means a_1 \cdot a_2 \cdot a_3 \cdots a_n.

Examples.

\prod_{k=1}^{5} k = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120 = 5!
\prod_{k=1}^{n} 2 = \underbrace{2 \cdot 2 \cdot 2 \cdots 2}_{n \text{ times}} = 2^n
\prod_{k=2}^{6} \frac{k}{k-1} = \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} = 6

That last product telescopes — most factors cancel, leaving \frac{6}{1} = 6. Telescoping in products works the same way as telescoping in sums: consecutive terms cancel.

Pi notation example: 5 factorialA large pi symbol on the left with limits k equals 1 to 5 and general term k. An arrow leads to the expanded product 1 times 2 times 3 times 4 times 5 equals 120. Π k=1 5 k 1 × 2 × 3 × 4 × 5 = 120 = 5! Pi notation compresses a multi-term product into a single expression.
Pi notation for $5! = 120$. The product runs from $k = 1$ to $k = 5$, multiplying each value of $k$.

Properties of pi notation.

\prod_{k=1}^{n} (c \cdot a_k) = c^n \cdot \prod_{k=1}^{n} a_k
\prod_{k=1}^{n} a_k \cdot b_k = \left(\prod_{k=1}^{n} a_k\right) \cdot \left(\prod_{k=1}^{n} b_k\right)

These follow from the commutativity and associativity of multiplication, just as the sigma rules follow from addition properties.

Telescoping sums

A telescoping sum is one where consecutive terms cancel, leaving only a few surviving terms — usually the first and the last. The name comes from a telescope: the sum collapses like the nested tubes of a collapsible telescope.

The standard telescoping identity

If you can write a_k = f(k) - f(k+1) for some function f, then:

\sum_{k=1}^{n} a_k = \sum_{k=1}^{n} [f(k) - f(k+1)] = f(1) - f(n+1)

The intermediate terms cancel: f(1) - f(2) + f(2) - f(3) + f(3) - f(4) + \dots + f(n) - f(n+1) = f(1) - f(n+1).

A classic example

Evaluate \sum_{k=1}^{99} \frac{1}{k(k+1)}.

Use partial fractions: \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}.

So:

\sum_{k=1}^{99} \frac{1}{k(k+1)} = \sum_{k=1}^{99} \left(\frac{1}{k} - \frac{1}{k+1}\right) = \frac{1}{1} - \frac{1}{100} = \frac{99}{100}

The sum of 99 fractions collapses to a single fraction. Without the telescoping trick, you would need to find a common denominator for 99 terms — essentially impossible by hand.

Telescoping cancellation in the sum of 1 over k times k plus 1A series of terms written as differences: (1 minus 1/2) plus (1/2 minus 1/3) plus (1/3 minus 1/4) plus dots plus (1/99 minus 1/100). Lines connect the negative part of each term to the positive part of the next term, showing they cancel. Only the first positive term 1 and the last negative term minus 1/100 survive. (1 − ½) + (½ − ⅓) + (⅓ − ¼) + ··· + (1/99 − 1/100) cancel cancel cancel 1 − 1/100 = 99/100 Everything in between cancels — only the first and last survive.
The telescoping mechanism: each $-\frac{1}{k+1}$ cancels the $+\frac{1}{k+1}$ from the next term. After all cancellations, only $1$ and $-\frac{1}{100}$ remain. The sum of $99$ fractions is $\frac{99}{100}$.

Recognising telescoping opportunities

The key step is the partial fraction decomposition (or a similar algebraic rewriting) that expresses the general term as a difference of consecutive values of some function. Common forms:

General term Rewrite as Telescopes to
\frac{1}{k(k+1)} \frac{1}{k} - \frac{1}{k+1} 1 - \frac{1}{n+1}
\frac{1}{k(k+2)} \frac{1}{2}\left(\frac{1}{k} - \frac{1}{k+2}\right) \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right)
\frac{2k+1}{k^2(k+1)^2} \frac{1}{k^2} - \frac{1}{(k+1)^2} 1 - \frac{1}{(n+1)^2}

The pattern: whenever the general term is a difference of a function evaluated at consecutive integers, the sum telescopes.

Worked examples

Example 1: Evaluate $\displaystyle\sum_{k=1}^{20} (2k - 1)$ using sigma properties and verify the result

Step 1. Split by linearity.

\sum_{k=1}^{20} (2k - 1) = 2\sum_{k=1}^{20} k - \sum_{k=1}^{20} 1

Why: the sum/difference rule and the constant-multiple rule let you separate the sum into two standard pieces.

Step 2. Evaluate each piece.

2\sum_{k=1}^{20} k = 2 \cdot \frac{20 \times 21}{2} = 420
\sum_{k=1}^{20} 1 = 20

Why: \sum k = \frac{n(n+1)}{2} and \sum 1 = n are the two most basic summation formulas.

Step 3. Combine.

\sum_{k=1}^{20} (2k - 1) = 420 - 20 = 400

Why: subtraction gives the final value.

Step 4. Verify: the sum 1 + 3 + 5 + \dots + 39 is the sum of the first 20 odd numbers. There is a well-known result: the sum of the first n odd numbers is n^2. Indeed, 20^2 = 400. Consistent.

Why: a known identity provides an independent check — if the two answers disagree, you know there is an error somewhere.

Result: \displaystyle\sum_{k=1}^{20} (2k - 1) = 400 = 20^2.

The sum of the first 20 odd numbers shown as a squareA 20 by 20 grid of small squares. The grid is divided into L-shaped gnomons: the first gnomon is 1 square in the corner, the second is an L-shape of 3 squares, the third is an L-shape of 5, and so on. Each gnomon is labelled with its size. The total grid has 400 squares. 1 3 5 20 × 20 = 400 1st odd: 1 2nd odd: 3 3rd odd: 5 ⋮ 20th odd: 39 Sum = 400 = 20²
The sum of the first $20$ odd numbers equals $20^2 = 400$. Each odd number $2k - 1$ forms an L-shaped gnomon around the previous square, extending it by one row and one column. The $20$ gnomons together build a $20 \times 20$ square.

The gnomon picture explains why the sum of odd numbers is a perfect square: each odd number wraps around the previous square to form a slightly larger square. The sigma notation \sum_{k=1}^{20} (2k-1) is the algebraic expression of this geometric fact.

Example 2: Evaluate $\displaystyle\sum_{k=1}^{50} \frac{1}{k(k+1)}$ using telescoping

Step 1. Decompose into partial fractions.

\frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}

Multiply both sides by k(k+1): 1 = A(k+1) + Bk. Set k = 0: A = 1. Set k = -1: B = -1.

\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}

Why: partial fractions rewrites the general term as a difference of consecutive terms — the signature of a telescoping sum.

Step 2. Write out the first few and last few terms.

\sum_{k=1}^{50} \left(\frac{1}{k} - \frac{1}{k+1}\right) = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{50} - \frac{1}{51}\right)

Why: writing out a few terms lets you see the cancellation pattern before jumping to the answer.

Step 3. Cancel.

Every -\frac{1}{k+1} from one term is killed by the +\frac{1}{k+1} in the next term. Only the very first positive part (\frac{1}{1}) and the very last negative part (-\frac{1}{51}) survive:

= 1 - \frac{1}{51} = \frac{50}{51}

Why: this is the telescope collapsing — 50 terms reduce to 2.

Step 4. Interpret.

The sum \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{50 \cdot 51} = \frac{50}{51} \approx 0.9804. As n increases, \sum_{k=1}^{n} \frac{1}{k(k+1)} = \frac{n}{n+1} \to 1. The infinite series converges to 1.

Why: the general formula \frac{n}{n+1} reveals the limiting behaviour — the sum approaches 1 but never reaches it for finite n.

Result: \displaystyle\sum_{k=1}^{50} \frac{1}{k(k+1)} = \frac{50}{51}.

Partial sums of the telescoping series approaching 1A graph showing the partial sums n over n plus 1 for n from 1 to 50. The points rise steeply at first and then level off approaching the horizontal asymptote y equals 1. The point at n equals 50 is highlighted at height 50 over 51. n Sₙ 0 10 20 30 40 50 1 y = 1 S₅₀ = 50/51
The partial sums $S_n = \frac{n}{n+1}$ rise steeply at first and then level off, approaching $1$ as $n \to \infty$. At $n = 50$, the sum is $\frac{50}{51} \approx 0.98$. The dashed line at $y = 1$ is the limit — the sum gets arbitrarily close but never reaches it for finite $n$.

The graph shows the diminishing returns of adding more terms: the first 10 terms get you to \frac{10}{11} \approx 0.91, but the next 40 terms only contribute about 0.07 more. The telescoping formula \frac{n}{n+1} captures this behaviour exactly.

Common confusions

Going deeper

If you came here to learn how sigma and pi notation work, manipulate sums using linearity, and evaluate telescoping sums, you have everything you need. The rest of this section is for readers who want to see more advanced uses and connections.

Index shifts and double sums

An index shift replaces k with k + c (or k - c) throughout the sum, adjusting the limits accordingly. This is the discrete analogue of substitution in integrals. For example:

\sum_{k=1}^{n} a_k = \sum_{k=0}^{n-1} a_{k+1}

The shift k \to k + 1 drops the lower limit by 1 and replaces a_k with a_{k+1}.

A double sum has two index variables:

\sum_{i=1}^{m} \sum_{j=1}^{n} a_{ij} = \sum_{i=1}^{m}\left(\sum_{j=1}^{n} a_{ij}\right)

The inner sum runs first (over j), producing a quantity that depends on i, and then the outer sum runs over i. When the limits are independent of each other, you can swap the order of summation: \sum_i \sum_j a_{ij} = \sum_j \sum_i a_{ij}. This is the discrete analogue of Fubini's theorem for double integrals.

Telescoping products

Telescoping works for products too. If a_k = \frac{f(k+1)}{f(k)}, then:

\prod_{k=1}^{n} a_k = \prod_{k=1}^{n} \frac{f(k+1)}{f(k)} = \frac{f(n+1)}{f(1)}

Everything in between cancels. The product \prod_{k=2}^{n} \frac{k}{k-1} = \frac{n}{1} = n is a simple instance. A more interesting example:

\prod_{k=2}^{n} \left(1 - \frac{1}{k^2}\right) = \prod_{k=2}^{n} \frac{(k-1)(k+1)}{k^2} = \prod_{k=2}^{n} \frac{k-1}{k} \cdot \prod_{k=2}^{n} \frac{k+1}{k}

The first product telescopes to \frac{1}{n} and the second to \frac{n+1}{2}, giving \frac{n+1}{2n}.

Connection to calculus

Sigma notation is the language of Riemann sums, which define the definite integral. The integral \int_a^b f(x) \, dx is defined as:

\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k^*) \Delta x

where \Delta x = \frac{b-a}{n} and x_k^* is a sample point in the k-th subinterval. Every skill you build with sigma notation — splitting sums, factoring constants, evaluating closed forms — transfers directly to computing integrals from the definition. The notation is not just shorthand; it is the bridge between algebra and calculus.

Interactive: telescoping partial sums of 1 over k(k+1)A graph of the partial sums of the telescoping series 1 over k times k plus 1. A draggable point controls the number of terms n. The readout shows n and the partial sum n over n plus 1. A dashed line at y equals 1 shows the limit. n Sₙ 0 10 20 30 1 limit drag to change n
Drag the point to watch the telescoping partial sum $S_n = \frac{n}{n+1}$ climb towards its limit of $1$. The first few terms contribute the most; later terms add less and less as the sum approaches but never quite reaches $1$.

Where this leads next