Sigma Summation Tool

Pick an expression, set bounds, and see how cumulative totals evolve term-by-term.

Worked example — Sigma Summation Tool

Finite sum	55
Term count	10

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Sigma Summation Tool

Sigma notation is a compact instruction to add. ∑ says: take the expression on its right, let the index run through every whole number from the lower limit up to the upper limit — both ends included — and add the results together.

The four parts of a sigma expression: the summation sign, with the upper limit written above it and the index and lower limit below, followed by the term to be added.

The summation sign. It carries no value — it tells you to add.
The upper limit n, above the sigma: the index's last value, included.
The index and its lower limit, k = m, below the sigma: where the counting starts, also included.
The term f(k), to the right: evaluated once per index value, and those results are what get added.

What each symbol means

k: the index — a counter that takes each whole-number value in turn. Mathematically its name is private to the sum, so i or j would mean exactly the same thing; here it is always k, and the table column above is headed Index (k).
m: the lower limit, written under the sigma as k = m. It is the index's first value, and it is included — it is the from field above.
n: the upper limit, written above the sigma. The last value the index takes, also included, which is why running from 1 to 10 gives ten terms rather than nine. It is the to field above.
f(k): the term — the expression evaluated once for each index value. Here it is chosen from five: k, k^2, 2k+1, 1/k and (-1)^k.

Where the formula comes from

Because both limits are included, the number of terms is not n − m. Write the index values out: m, m + 1, and so on up to n.
Subtract m from every entry so the list starts at zero: 0, 1, … , n − m. Relabelling changes what the entries are called, not how many there are.
A list running from 0 to n − m holds n − m + 1 entries. So 1 up to 10 gives ten terms and 0 up to 12 gives thirteen — which is what Term count reports above in each case.

Assumes: Whole-number index values, one step at a time, and finitely many of them: both limits are rounded to integers and clamped to the range −30 to 30 so the page cannot lock up. The term is picked from five presets rather than typed, which makes this a demonstrator for the notation rather than a general-purpose summation calculator.
Breaks when: Choose 1/k and start the index at 0, and the first term is undefined. It is not quietly skipped — the calculation stops there: the result reads undefined, Term count falls to 1, and the breakdown table ends at that single row. The written-out expansion still shows undef + 1/1 + 1/2 + ⋯, which is the honest picture — one undefined term makes the whole sum undefined, no matter how many well-behaved terms follow it.

Every running total in the odd-number preset is a perfect square 🖖

Load the 2k+1 preset and read the cumulative column downwards: 1, 4, 9, 16, 25 — it never stops being a square, and it ends on 169, which is 13². That is not a coincidence of these bounds. Adding the first n odd numbers always gives exactly n², so the running total the tool prints after each step is the count of terms so far, squared. The partial sums are doing geometry: each odd number is the L-shaped strip that turns one square into the next.

Shorthand for adding things up 🖖

The big Σ is simply an instruction: add up every term while the index climbs from the lower bound to the upper bound. This tool does the adding for you and, just as usefully, shows the running total after each step — the partial sums. The final answer is nothing more than the last partial sum in that trajectory.

When adding nothing gives zero 🖖

An empty sum — one with no terms at all, because the upper limit sits below the lower — is not undefined. By convention it equals 0: zero is the additive identity, so an empty total leaves whatever it is added to unchanged. Its multiplicative twin is just as surprising, an empty product being 1. You cannot make one here, though, and that is worth knowing: set an upper limit below the lower one and the two are silently swapped, so k = 5 down to 2 becomes k = 2 up to 5 and returns 14.

Euler wrote it down, and you already click it 🖖

The symbol is Euler’s. In Institutiones calculi differentialis (1755) he set down that a sum would be marked by the sign Σ — Greek capital sigma, the initial of summa — and it stuck, which for mathematical notation is rare: most proposals from that era were replaced. It has long since left mathematics behind. The AutoSum button in every major spreadsheet is that same Σ, so people who would never describe themselves as mathematicians reach for Euler’s notation most working days. One caution, because the two look related and are not: capital Σ is an instruction, add these up, while lowercase σ is the standard deviation of a distribution. “Six Sigma” is the lowercase one.

Practice

Check yourself

Predict the answer first, then use the controls above to find out. Reveal only after you have committed to a guess — that is what makes it practice.

The odd numbers preset sums 2k+1 from k = 0 to 12 and reports 169 over 13 terms. The other presets print a closed form underneath their total; this one prints none. Find the pattern yourself — set the upper limit to 4, then 9, then 19.

Show answer

25, 100 and 400, over 5, 10 and 20 terms. The sum of the first n odd numbers is exactly n², every time, and 169 is 13². Picture a square built in L-shaped layers: one dot, then a three-dot L wrapped round it, then a five-dot L round that — after n layers you are holding an n × n square, and each layer you added was the next odd number. The tool recognises the closed forms for k and k² and prints them; it does not recognise this one, so the total is all you get.
The harmonic preset sums 1/k from 1 to 8 and gets 2.717857, with a note saying the partial sums never settle on a finite limit. Guess how far the upper limit has to go before the sum passes 5 — then find out.

Show answer

83. Term 82 leaves it at 4.990020 and term 83 tips it to 5.002068. Reaching 6 takes 227 terms, and 226 lands on 5.999961 — four millionths short. Notice 227 ÷ 83 = 2.73, which is very nearly e: the partial sums track ln n, so every further step of +1 costs about e ≈ 2.718 times as many terms as everything before it. Passing 10 takes 12367 terms; the tool will do it if you ask. That is what "never settles" looks like from the inside — it never stops climbing, but it slows down geometrically, which is why eight terms only got you to 2.7.

Problem solved in full

The sum of squares which pairing 1 to 10 cannot reach 6 steps
Sum 1 to 10 by pairing rather than by adding, then use the same trick to get the sum of squares — which pairing cannot reach.
1. Write the sum out. Ten terms is small enough to add directly, which is exactly why it is the wrong thing to do.
  
  $S = k = 1 \sum 10 k = 1 + 2 + \dots + 10$
2. Add the sum to its own reverse. Every column gives the same total, and there are n of them — so the doubled sum is a rectangle.
  
  $2 S = (1 + 10) + (2 + 9) + \dots + (10 + 1) = 10 \times 11 ⟹ S = \frac{10 \cdot 11}{2} = 55$
3. The general result. Note the argument doubles the sum rather than folding it in half, which is why it needs no separate case for odd n — the version that pairs terms from both ends does.
  
  $k = 1 \sum n k = \frac{n ( n + 1 )}{2} — the pairing needs no case split, because it doubles the sum rather than folding it$
4. Now squares. Start from a difference of consecutive cubes, which expands to something containing k².
  
  $(k + 1)^{3} - k^{3} = 3 k^{2} + 3 k + 1 ⟹ k = 1 \sum n [(k + 1)^{3} - k^{3}] = (n + 1)^{3} - 1$
5. Sum both sides. The left telescopes to a single term; the right contains the sum you want plus the sum you already have.
  
  $(n + 1)^{3} - 1 = 3 \sum k^{2} + 3 \cdot \frac{n ( n + 1 )}{2} + n ⟹ k = 1 \sum n k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6} = 385$
6. One more power and the pattern closes on itself.
  
  $k = 1 \sum n k^{3} = (\frac{n ( n + 1 )}{2})^{2} = 5 5^{2} = 3025$
Answer

55, then 385, then 3025 — and the third one is the first one squared. The pairing argument is famous and it is also a dead end: it works because the terms are linear, and there is no way to fold squares into equal pairs. What replaces it is telescoping. Write down a difference of consecutive cubes, sum it, and almost everything cancels; the sum of squares falls out of the wreckage. The same machine run one power higher gives the sum of cubes, and it lands on 55² — the sum of the first n cubes is the square of the sum of the first n numbers, for every n. That is not a coincidence you can see by staring at the numbers, and it is why telescoping is worth learning even though pairing gets the first answer faster.

$k = 1 \sum 10 k = 55, \sum k^{2} = 385, \sum k^{3} = 5 5^{2} = 3025$

References (3)

Where the Σ sign comes from — Euler’s own text: L. Euler, Institutiones calculi differentialis cum eius usu in analysi finitorum ac doctrina serierum, vol. 1. Academiae Imperialis Scientiarum Petropolitanae, 1755 (E212).
And the history of the notation itself: Florian Cajori, A History of Mathematical Notations, Volume II. The Open Court Publishing Company, 1929.
The conventions this page relies on — inclusive limits, index changes, and why an empty sum is 0: R. L. Graham, D. E. Knuth & O. Patashnik, Concrete Mathematics, 2nd ed., chapter 2 (“Sums”). Addison-Wesley, 1994. ISBN 978-0-201-55802-9.

By the 42Math team · Updated 2026-05-21

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

All Mathematics lessons →

Example problems

natural numbers - Classic sum of first ten natural numbers.
squares - Finite sum of squares with rapidly increasing terms.
linear expression - Linear-term accumulation with odd-number pattern.
harmonic partial sum - Partial harmonic sum demonstrates slow logarithmic growth.