Some special sums


Some notation

In this course, we will be considering lots of different sums and it will be useful to have some notation for representing them compactly. For instance, in the previous example, we needed to consider the sum

\[ 1 + 2 + 3 + 4 + \ldots + k \]

We will represent this using the so-called sigma notation

\[ \sum_{i=1}^k i \]

To decode this, the expressions on the bottom and top of the Greek letter sigma give us a range for a quantity which is denoted by $ i $ . In this case, we are told that i varies between 1 and k . What follows the sigma tells us what we are going to add. Here we are being told simply to add i . In other words, this expression means to add i as i varies between 1 and k .

Here are a few more examples:

\begin{eqnarray*} \sum_{i=1}^4 i^2 & = & 1^1 + 2^2 + 3^2 + 4^2 \\ \sum_{i=0}^{10} \cos(i \pi) & = & \cos(0) + \cos(\pi) + \cos(2\pi) +\ldots + \cos(10\pi) \end{eqnarray*}


Summing an arithmetic Sequence

The first series of interest to us is the so-called arithmetic sequence,

\[ S = 1 + 2 + 3 + 4 + \ldots + k = \sum_{i=1}^k i \]
Below, we will show two different approaches to summing this finite list of the first k integers. The first method involves a cute trick. We will write the sum twice, as follows:

\[ \begin{array}{ccccccccccccc} S & = & 1 & + & 2 & + & 3 & + & 4 & + & \ldots & + & k \\ S & = & k & + & k-1 & + & k-2 & + & k-3 & + & \ldots & + & 1 \\ \end{array} \]

Now add the two rows together.

\[ 2S = (k+1) + (k+1) + (k+1) + \ldots + (k+1) \]

Notice that each of the columns on the right hand side equals $ k + 1 $ . And how many of these columns are there? k, of course. This means that $ 2S = k(k+1) $ or

\[ S = \frac{k(k+1)}{2} \]


It is well known that when Gauss, who would later become a great mathematician, was just 10 years old, his teacher assigned what was to be a long and time-consuming excercise to his school class: to add up all the integers from 1 to 100 . This was supposed to keep the class occupied for the morning, but much to the chagrin of his teacher, Gauss finished the task immediately. He had already discovered the above summation formula by using the same trick that we show above. (Incidentally, we are told that he had the only correct answer, since his friends who added up term by term made many mistakes in the process.)

Since this time, the formula we have developed has come to be named after Gauss. It is amazingly useful, as we shall soon see.

Even though the above proof is a very elegant way of establishing Gauss's formula, we will spend a bit more time on it below, and show yet one more method of arriving at the same result. The reason: we can then use this new method (which at first sight seems a bit more complicated) to sum other series in which we are also interested.


Proof # 2 of Gauss's Formula for the sum of integers

Below we show a second proof of Gauss's formula

\[ \sum_{i=1}^{k} ~~i = \frac{k(k+1)}{2} \]




Proof:

Start with the following fact:

\[ (n+1)^2=n^2 + 2n + 1 \]

This implies that

\[ (n+1)^2-n^2 = 2n + 1 \]

Now write the following list, one for each value of n from n=1 to n=k.

\[2^2 -1^2 = 2 \times 1 + 1 \] \[ 3^2 -2^2 = 2 \times 2 + 1\] \[ 4^2 -3^2 = 2 \times 3 + 1 \] \[ 5^2 -4^2 = 2 \times 4 + 1\] \[ \vdots \] \[ (k+1)^2 -k^2 = 2 \times k + 1\]

Add both sides of the above list, noticing that many of the terms on the left hand side cancel out. The only terms that do not cancel from the left hand side are $ 1^2=1 $ and $ (k+1)^2 $ . On the other side, there are many 1's (how many of them?) and 2 multiplied by the sum (1 + 2 + 3+..). When we add all these equations together, we obtain:

\[ (k+1)^2 -1 = 2 (1+2+3+ \dots + k) + k \times 1 \]

Rearranging this result leads to

\[ \frac{(k+1)^2 -1 - k}{2} =(1+2+3+ \dots + k) \]

We can simplify the expression on the left, and notice that this is the result we were looking for, namely the sum of the first k integers. We obtain

\[ \frac{(k+1)^2 -1 - k}{2} = \frac{(k+1)^2 -(k+1)}{2} = \frac{(k+1)(k+1-1)}{2} = \frac{(k+1)k}{2} = \sum_{i=1}^{k} i \]

We have thus arrived at the required result.




Other, related, summation formulae

One of the attractive features of the proof shown above is that it can be generalized to more interesting and involved sums, including each of the following cases.

The sum of squares

\[ \sum_{n=1}^{k} ~~n^2 = \frac{k(k+1)(2k+1)}{6} \]



The sum of cubes
\[ \sum_{n=1}^{k} ~~n^3 = \left(\frac{k(k+1)}{2}\right)^2 \]

Comment: to establish the formula for the sum of squares, we would start out with the cubic equation

\[ (n+1)^3 = n^3 + 3n^2+3n+1 \]

and go through a very similar procedure to the one we used for Gauss's formula. Similarly, to establish the formula for sums of cubes, we would use as a starting point, the quartic equation, etc. This is left as an exercise (see below) for the reader.



Examples and calculations involving sums

Summation notation can be confusing at first, but after getting used to it, it will become familiar. Here we illustrate a few examples involving computation of sums.

(a) Compute the sum
\[ \sum_{n=10}^{20} ~~(n^2 + 1 ) \]
We can reduce this to a more familiar problem, by noticing that since these are sums, an equivalent way to write this is:

\[ \sum_{n=10}^{20} ~~(n^2 + 1) = \sum_{n=10}^{20} ~~(n^2) + \sum_{n=10}^{20}1~~ = \sum_{n=1}^{20} ~~(n^2) -\sum_{n=1}^{9} ~~(n^2) +11 (1) \]

(Note that the sum of the 1's is 11, since we are taking the sum from n=10 to n=20.) We can now apply the formula for sums of squares to each of the above pieces, arriving at:

\[ \sum_{n=10}^{20} ~~(n^2 + 1 ) = \frac{(20)(21)(41)}{6} - \frac{(9)(10)(19)}{6} + 11 = 2870 - 285 +11 = 2596 \]


For your consideration: