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Calculus 2

Unit 1: Lesson 3

Summation notation review

Summation notation

We can describe sums with multiple terms using the sigma operator, Σ. Learn how to evaluate sums written this way.
Summation notation (or sigma notation) allows us to write a long sum in a single expression.

Unpacking the meaning of summation notation

This is the sigma symbol: sum. It tells us that we are summing something.
\begin{aligned} \scriptsize\text{Stop at }n=3& \\ \scriptsize\text{(inclusive)} \\ \searrow\qquad& \\\\ \LARGE\displaystyle\sum_{n=1}^3&\LARGE 2n-1 \\ &\qquad\quad\nwarrow \\ \nearrow\qquad&\qquad\scriptsize\text{Expression for each} \\ \scriptsize\text{Start at }n=1&\qquad\scriptsize\text{term in the sum} \end{aligned}
This is a summation of the expression 2, n, minus, 1 for integer values of n from 1 to 3:
\begin{aligned} &\phantom{=}\displaystyle\sum_{\goldD n=1}^3 2\goldD n-1 \\\\ &=\underbrace{[2(\goldD 1)-1]}_{\goldD{n=1}}+\underbrace{[2(\goldD 2)-1]}_{\goldD{n=2}}+\underbrace{[2(\goldD 3)-1]}_{\goldD{n=3}} \\\\ &=1+3+5 \\\\ &=9 \end{aligned}
Notice how we substituted start color #e07d10, n, equals, 1, end color #e07d10, start color #e07d10, n, equals, 2, end color #e07d10, and start color #e07d10, n, equals, 3, end color #e07d10 into 2, start color #e07d10, n, end color #e07d10, minus, 1 and summed the resulting terms.
n is our summation index. When we evaluate a summation expression, we keep substituting different values for our index.
Problem 1
sum, start subscript, n, equals, 1, end subscript, start superscript, 4, end superscript, n, squared, equals, question mark

We can start and end the summation at any value of n. For example, this sum takes integer values of n from 4 to 6:
\begin{aligned} &\phantom{=}\displaystyle\sum_{\goldD n=4}^6 \goldD n-1 \\\\ &=\underbrace{(\goldD 4-1)}_{\goldD{n=4}}+\underbrace{(\goldD 5-1)}_{\goldD{n=5}}+\underbrace{(\goldD 6-1)}_{\goldD{n=6}} \\\\ &=3+4+5 \\\\ &=12 \end{aligned}
We can use any letter we want for our index. For example, this expression has i for its index:
\begin{aligned} &\phantom{=}\displaystyle\sum_{\goldD i=0}^2 3\goldD i-5 \\\\ &=\underbrace{[3(\goldD 0)\!-\!5]}_{\goldD{i=0}}+\underbrace{[3(\goldD 1)\!-\!5]}_{\goldD{i=1}}+\underbrace{[3(\goldD 2)\!-\!5]}_{\goldD{i=2}} \\\\ &=-5+(-2)+1 \\\\ &=-6 \end{aligned}
Problem 2
sum, start subscript, k, equals, 3, end subscript, start superscript, 5, end superscript, k, left parenthesis, k, plus, 1, right parenthesis, equals

Problem 3
Consider the sum 4, plus, 25, plus, 64, plus, 121.
Which expression is equal to the above sum?

Some summation expressions have variables other than the index. Consider this sum:
sum, start subscript, n, equals, 1, end subscript, start superscript, 4, end superscript, start fraction, k, divided by, n, plus, 1, end fraction.
Notice that our index is n, not k. This means we substitute the values into n, and k remains unknown:
\begin{aligned} &\phantom{=}\displaystyle\sum_{\goldD n=1}^3 \dfrac{k}{\goldD n+1} \\\\ &= \dfrac{k}{(\goldD 1)+1} + \dfrac{k}{(\goldD 2)+1} + \dfrac{k}{(\goldD 3)+1} \\\\ &= \dfrac{k}{2} + \dfrac{k}{3} + \dfrac{k}{4} \end{aligned}
Key takeaway: Before evaluating a sum in summation notation, always make sure you identified the index, and that you are only substituting into that index. Other unknowns should remain as they are.
Problem 4
sum, start subscript, m, equals, 1, end subscript, start superscript, 4, end superscript, 8, k, minus, 6, m, equals, question mark

Want more practice? Try this exercise.

Want to join the conversation?

• How do i derive the formula for summation?

Sum from k to n i = [(n-k+1)(n+k)]/2
• Another way to derive this formula is to let
S = Sum from k to n of i, write this sum in two ways, add the equations, and finally divide both sides by 2. We have

S = k + (k+1) + ... + (n-1) + n
S = n + (n-1) + ... + (k+1) + k.

When we add these equations, we get 2S on the left side, and n-k+1 column sums that are each n+k on the right side.

So 2S = (n-k+1)(n+k).
Dividing both sides by 2 gives S = [(n-k+1)(n+k)]/2.
• In my physics class the derivative of momentum was taken and the summation went from having k=1 on the bottom and N on the top to just k on the bottom, why is this? Is it the same thing, but short hand?
• In the first section (Unpacking Sigma Notation), I've seen the index equal 0. But my calculus teacher says that the index can't be 0, because you can't have the 0th term of a sequence. But all else being equal (the sequence and summation index remaining the same), what would be the difference between a sum with i = 0 and a sum with i = 1?

Thank you.
• Nothing really. Nothing changes if you shift all the indices down by 1. In fact, you can really start at any index you want because there's no convention that the subscript has to denote which number the term is in the sequence. Generally, people start at index 1 because it happens to be convenient to use the subscripts (and so the indices) to keep track of the number of the terms.
• How about when your "stopping point" number is super high, to a point where writing out all of the sums would take ages? How can we shorten it or plug it into another equation?
• Either you would find a formula that depends on the upper limit (like with a finite geometric series), use methods of cancellation (like with a telescoping series), or just numerically compute it (using a calculator).
• How do I solve for the number on top of the Sigma?

If I know the starting index, and I know the formula and the final sum, how do I solve for the ending index?

Example, how do I solve for X?
X
E f(n * 500) = 18000
n = 1
• Why is (sigma; n=1 to 3) (n!) -n undefined?
• I have no idea, it doesn't look like it should be undefined. If we just calculate the sum from 1 to 3, we get a perfectly defined number:
Sum = (1! - 1) + (2! - 2) + (3! - 3)
= 0 + 0 + 3 = 3
If you set n to infinity though, the series will diverge and there will be no sum. Could you be confusing it with that?
• Additionally, does the "i" simply go up by integer values? What if I wanted a different manner of arithmetically changing "i"?
• Some books/teachers will accept writing, for example, 'step 0.5' to indicate that i is going up in increments of 1/2.
However, this isn't necessary, since you can just change the expression inside the sum to get the same effect.

So if you have the sum from i=1 to 10 of i², and you want to i to step by 1/3 instead of 1, you can change the expression i² to (1+(i-1)/3)², and have i go from i=1 to 28. This will get you the same sequence of numbers as just changing the step.
• What if my index is m and is in the exponent of some "not index" variable?