Summation Notation

Summation notation is used to denote a sum of terms. Usually, the terms follow a pattern or formula.

$$\sum_{k=1}^n f(k) \quad\hbox{is shorthand for}\quad f(0) + f(1) + \cdots + f(n).$$

In this case, 1 and n are the limits of summation and k is the summation variable. Each integer from the lower limit (in this case, 1) to the upper limit (in this case, n), including both limits, is substituted in turn for each occurence of k in the term $f(k)$ . After that, the results are added.

You may use any variable for the summation variable, though most often people use the letters from i through n. The lower and upper limits may be any integers, as long as the upper limit is greater than or equal to the lower limit. The term in the sum may be any function of the summation variable; the summation variable may also be used as a subscript of superscript.


Example.

$$\sum_{n=1}^5 n^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55.$$

$$\sum_{n=1}^4 7 = 7 + 7 + 7 + 7 = 28.$$

$$\sum_{j=-2}^2 \sqrt{b_j} = \sqrt{b_{-2}} + \sqrt{b_{-1}} + \sqrt{b_0} + \sqrt{b_1} + \sqrt{b_2}.$$

$$\sum_{k=0}^3 \dfrac{1}{k + 1} = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4}.$$

$$\sum_{i=0}^\infty a_i = a_0 + a_1 + a_2 + a_3 + \ldots.$$

It's often useful to go from a sum of terms written out "the long way" to a summation:

$$\sin (-2) + \sin (-1) + \sin 0 + \sin 1 + \sin 2 + \sin 3 = \sum_{n=-2}^3 \sin n.$$

Notice that you can also write

$$\sin (-2) + \sin (-1) + \sin 0 + \sin 1 + \sin 2 + \sin 3 = \sum_{n=-1}^4 \sin (n - 1).$$

There are infinitely many ways to write a sum using summation notation. It's a good habit to check that you have the correct range for your index by plugging in the top and bottom numbers. For example, using $\displaystyle
   \sum_{n=-1}^4 \sin (n - 1)$ , I find that

$$n = -1 \quad\hbox{gives}\quad \sin (-1 - 1) = \sin (-2),$$

$$n = 4 \quad\hbox{gives}\quad \sin (4 - 1) = \sin 3.$$

These are the first and last terms for the original sum, so I have some assurance that I set it up correctly.


Example. In some cases, you can use your calculator or a computer to approximate a sum. For example, I computed the following sum using a computer:

$$\sum_{n=1}^{50} \dfrac{1}{n} \approx 4.49921.\quad\halmos$$


Sometimes you can obtain a formula for a sum. For example,

$$\sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \dfrac{n(n + 1)}{2},$$

$$\sum_{k=1}^n k^2 = 1^2 + 2^2 + 3^2 + \cdots + n^2 = \dfrac{(2n + 1)n(n + 1)}{6},$$

$$\sum_{k=0}^n r^k = \dfrac{1 - r^{n+1}}{1 - r}.$$

Here's a pictorial argument for the first formula:

$$\hbox{\epsfysize=2in \epsffile{sums1.eps}}$$


Example. Evaluate the following sums:

(a) $\displaystyle \sum_{k=1}^{100} k$ .

$$\sum_{k=1}^{100} k = 1 + 2 + 3 + \cdots + 100 = \dfrac{100\cdot 101}{2} = 5050.\quad\halmos$$

(b) $\displaystyle \sum_{k=1}^{100} k^2$ .

$$\sum_{k=1}^{100} k^2 = 1^2 + 2^2 + 3^2 + \cdots + 100^2 = \dfrac{201\cdot 100\cdot 101}{6} = 338350.\quad\halmos$$

(c) $\displaystyle \sum_{k=0}^{100}
   \left(\dfrac{1}{2}\right)^k$ .

$$\sum_{k=0}^{100} \left(\dfrac{1}{2}\right)^k = \dfrac{1 - \left(\dfrac{1}{2}\right)^{101}}{1 - \dfrac{1}{2}} = 2 - \left(\dfrac{1}{2}\right)^{100}.\quad\halmos$$


Example. Find a closed-form expression for $\displaystyle \sum_{k=1}^n (3k^2 + 5k +
   4)$ .

$$\sum_{k=1}^n (3k^2 + 5k + 4) = 3 \sum_{k=1}^n k^2 + 5 \sum_{k=1}^n k + \sum_{k=1}^n 4 = 3\cdot \dfrac{(2n + 1)n(n + 1)}{6} + 5\cdot \dfrac{n(n + 1)}{2} + 7n.\quad\halmos$$


Example. (Telescoping series) Use the fact that

$$\dfrac{2}{k^2 + 2k} = \dfrac{1}{k} - \dfrac{1}{k + 2}$$

to compute $\displaystyle
   \sum_{k=1}^{1000} \dfrac{2}{k^2 + 2k}$ .

$$\sum_{k=1}^{1000} \dfrac{2}{k^2 + 2k} = \sum_{k=1}^{1000} \left(\dfrac{1}{k} - \dfrac{1}{k + 2}\right) =$$

$$\left(\dfrac{1}{1} - \dfrac{1}{3}\right) + \left(\dfrac{1}{2} - \dfrac{1}{4}\right) + \left(\dfrac{1}{3} - \dfrac{1}{5}\right) + \left(\dfrac{1}{4} - \dfrac{1}{6}\right) + \cdots + \left(\dfrac{1}{999} - \dfrac{1}{1001}\right) + \left(\dfrac{1}{1000} - \dfrac{1}{1002}\right) =$$

$$\dfrac{1}{1} + \dfrac{1}{2} - \dfrac{1}{1001} - \dfrac{1}{1002} = \dfrac{751250}{501501}.$$

I used the given identity to rewrite the sum, then I wrote out enough of the terms that I could see what's happening. All the terms except for the four in the last line cancel: Each term from $\dfrac{1}{3}$ to $\dfrac{1}{1000}$ appears once with positive sign and once with negative sign. This leaves the four terms in the last line.


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