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Section 7.2 Summation notation

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Subsection 7.2.1 Basic definitions

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Summation notation is a compact form for writing sums. It is written with three pieces. ∑k=152k The part below the greek letter sigma (Σ) tells us to start with k=1. We increment k by 1 until we get to the value above the sigma. In this case we let k take on the values 1,2,3,4,5. We next substitute those values of k into the expression to the right of the sigma successively and take the sum all of the resulting terms. Here is a table of the resulting values:

k2k12243648510

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which evaluates to

2+4+6+8+10=30

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and so we write

∑k=152k=30.

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The variable k is called the index of summation. The number above the sigma is called the limit of summation. The example shows us how to write a sum of even numbers. We can write the sum of odd numbers, too.

∑k=18(2k−1)=1+3+5+7+9+11+13+15=64.

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It is not necessary to use k as a variable or to start the summation at k=1. It is easy to verify that

∑k=18(2k−1)=∑l=07(2l+1)=64.

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The number above the sigma can also be written as a variable:

∑k=1nak=a1+a2+⋯+an−1+an∑k=0nakbk=a0b0+a1b1+a2b2+⋯+an−1bn−1+anbn∑k=1n(2k−1)=1+3+5+7+⋯+(2n−1)

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Next we take the last expression and vary n, the limit of summation. Here is a table that includes some small values of n:

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Table 7.2.1.
∑k=11(2k−1)=1 n=1
∑k=12(2k−1)=4 n=2
∑k=13(2k−1)=9 n=3
∑k=14(2k−1)=16 n=4
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∑k=1n(2k−1)=1+3+5+⋯+2n−1=?

What simple expression involving \(n\) can replace the question mark?

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The summands can be symbolic, too. If f(x) is any function, we may write

∑i=1rf(i)=f(1)+f(2)+⋯+f(r−1)+f(r)
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Subsection 7.2.2 Alternating sums

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A sum is alternating if the signs of consecutive terms are alternate between positive and negative. An example: 1−2+3−4+5. These sums appear frequently enough to have this special name. They are easy to write using summation notation because of a simple fact:

(−1)k={1if k is even−1if k is odd

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We can then write

∑k=16(−1)kk2=−1+4−9+16−25+36=21.

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We can interchange the order of the signs by adding (or subtracting) one to the exponent of −1:

∑k=16(−1)k+1k2=1−4+9−16+25−36=−21.
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Subsection 7.2.3 Manipulation with summation notation identities

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Summation notation is more that mere shorthand. There are rules of manipulation that are quite useful.

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For any constant c,

∑k=1nc=c+c+⋯+c+c⏟n summands=nc

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and

∑k=1ncak=ca1+⋯+can=c(a1+⋯+an)=c∑k=1nak

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The process of the last equation is sometimes referred to as pulling a constant across the summation sign.

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In addition,

∑k=1n(ak+bk)=∑k=1nak+∑k=1nbk,∑k=1n(ak−bk)=∑k=1nak−∑k=1nbk

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and

∑k=0nak=∑k=tn+tak−t

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The last identity is sometimes called shifting the index of summation.

Suppose we want to evaluate the sum of the first \(n\) positive integers. In other words, we want to find

\begin{equation*} s_n=1+2+\cdots+n=\sum_{k=1}^nk. \end{equation*}

If we are convinced that \(\sum_{k=1}^n(2k-1)=1+3+5+\cdots+2n-1=n^2\text{,}\) then

\begin{align*} n^2\amp =\sum_{k=1}^n(2k-1)\\ \amp =\sum_{k=1}^n(2k)-\sum_{k=1}^n1\\ \amp =2\sum_{k=1}^nk-n\\ \amp =2s_n-n \end{align*}

which yields

\begin{equation*} s_n=\frac12n(n+1). \end{equation*}

Evaluate the following sums

  • \(\displaystyle \sum_{j=1}^5 2j\)

  • \(\displaystyle \sum_{k=1}^5 k(k-1)\)

  • \(\displaystyle \sum_{i=0}^4 2^i\)

  • \(\displaystyle \sum_{\ell=1}^3\ell^2+\frac1\ell\)

  • \(\displaystyle \sum_{j=1}^5 ij\)

Solution.

  • \(\displaystyle 2+4+6+8+10=30\)

  • \(\displaystyle 2+6+12+20+30=70\)

  • \(\displaystyle 1+2+4+8+16=31\)

  • \(\displaystyle 1+1+4+\frac12+9+\frac13=15\frac56\)

  • \(\displaystyle i+2i+3i+4i+5i=15i\)

Evaluate \(\sum_{k=0}^n 2^k\) for \(n=1,2,3,4\text{.}\) Find a simple expression that works for any \(n\text{.}\) Justify your answer.

Solution.
\begin{equation*} \begin{array}{cc} n \amp \sum_{k=0}^n 2^k\\ \hline 1 \amp 3\\ 2 \amp 7\\ 3\amp 15\\ 4\amp 31 \end{array} \end{equation*}

We note that each answer is one less than a power of \(2\text{.}\) This leads to the equation \[ \sum_{k=0}^n 2^k=2^{n+1}-1. \] To justify the equation we note that \[ 2(\sum_{k=0}^n 2^k)=\sum_{k=0}^n 2^{k+1}=\sum_{k=1}^{n+1} 2^k =\bigl(\sum_{k=0}^{n+1} 2^k\bigr) -1 \] so we get the sum for a given value on \(n\) by doubling the value for \(n-1\) and adding \(1\text{.}\)

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Subsection 7.2.4 Double summations

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Sometimes it is useful to have a formula with two summations, each with its own index of summation. It is interpreted in the following way: ∑i=1m∑j=1nf(i,j)=∑i=1m(∑j=1nf(i,j)) and so, for example, ∑i=13∑j=12ij=∑i=13(i1+i2)=(1+1)+(2+4)+(3+9)=20

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Double summations are often used with matrices. Suppose we have a matrix

A=[a1,1a1,2⋯a1,na2,1a2,2⋯a2,n⋮⋮⋮⋮am,1am,2⋯am,n]

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Then ∑j=1nai,j=ai,1+ai,2+⋯+ai,n which is the sum of the entries in the i-th row of the matrix. Similarly ∑i=1mai,j=a1,j+a2,j+⋯+am,j which is the sum of the entries in the j-th column of the matrix. We can now see that

∑i=1m∑j=1nai,j=∑j=1n∑i=1mai,j

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since each sides of the equation is equal to the sum of all of the entries in the matrix. This is called interchanging the order of summation.

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Another way of looking at these double summations ∑i=1m∑j=1nf(i,j) is that it is the sum of f(i,j) as i takes on values from 1 to m and j takes on values from 1 to n. Clearly there are mn summands altogether.