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Arithmetic Series

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If you are building a football stadium and each row has 10 fewer seats as you go toward the bottom of the stadium, how do you figure out how many seats will be in the stadium? The answer lies in arithmetic series.

First, remember that an arithmetic sequence is one where the distance between terms is a constant. So an example of an arithmetic sequence would be \( \{ 3, 6, 7, 9, \dots \} \), where the distance between successive terms is always \( 2 \).

In general, an **arithmetic sequence** has the form \( \{a, a+d, a+2d, a+3d, \dots \} \) where \(a \) and \( d \) are real numbers. The distance between successive terms is always \( d\).

An **arithmetic series** is one where the terms of the series are an arithmetic sequence.

Naturally, you want a compact way to write out an arithmetic series.

An **arithmetic series** is a series of the form

\[ \sum\limits_{n=0}^\infty (a + dn) \]

where \(a \) and \( d \) are constant real numbers.

You might wonder if an arithmetic series converges. The first thing to try is the n^{th} Term Test for Divergence. The n^{th} Term Test for Divergence tells you that if

\[ \lim\limits_{n \to \infty} (a+dn) \]

does not exist, or has a limit which isn't zero, then the arithmetic series diverges. But \( a \) and \( d\) are constants, so if \( d \not= 0\) this limit doesn't exist and the arithmetic series diverges. On the other hand, if \(d=0 \) then

$$ \lim\limits_{n \to \infty} (a+dn) = \lim\limits_{n \to \infty} a = a $$

Since \( a \not= 0\) then you know that the arithmetic series diverges. That leads you to exactly one arithmetic series being able to converge, and that is the one where the corresponding arithmetic sequence is \( \{0, 0, 0, \dots \} \), which is not very interesting!

So then, why do you care about arithmetic series at all?

The answer to why you might care about arithmetic series lies in the partial sums. Even if the series doesn't converge, the partial sums can still be useful. So let's examine the partial sums for the arithmetic series:

$$ \begin{array}{lll} s_0 &=& a \\ s_1 &=& 2a + d \\ s_2 &=& 3a + 3d \\ s_3 &=& 4a + 6d \\ s_4 &=& 5a + 10d. \end{array} $$

This isn't looking very helpful so far in trying to find a pattern. Sometimes in math, you try something and then find out it wasn't very helpful, so you try something different. This is part of the discovery process and doesn't mean you did something wrong!

Instead let's try looking at the whole partial sum \( s_{n-1} \) written out in longer form:

\[ s_{n-1} = a + (a + d) + \dots + (a + (n-2)d) + (a + (n-1)d) . \]

It looks like many of those terms are similar, so what happens if you write them backward? Then

\[ s_{n-1} = (a + (n-1)d) + (a + (n-2)d) + \dots + (a + d) + a . \]

If you add the two of those together things might cancel out, so giving that a try:

\[ \begin{array}{lll} 2s_{n-1} &=& a + (a + d) + \dots + (a + (n-2)d) + (a + (n-1)d) \\ && + (a + (n-1)d) + (a + (n-2)d) + \dots + (a + d) + a \\ &=& [a + (a + (n-1)d] + [(a + d) +(a + (n-2)d) ] + \dots \\ && + [ (a + (n-2)d) + (a + d) ] + [(a + (n-1)d) + a] \\ &=& n[2a + (n-1)d] \end{array} \]

because there are exactly \(n \) repeats of \( 2a + (n-1)d \). So it was a strange thing to try, but it certainly seems to have helped! Now all that is left is to divide by 2 to get

\[ s_{n-1} = \frac{n}{2} (2a + (n-1)d) . \]

Sometimes you will see this written as

\[ \sum\limits_{k=0}^{n-1} (a+kd) = \frac{n}{2} (2a + (n-1)d) . \]

There is more than one way to think of the same sum. Suppose you want to add up the natural numbers 1, 2, 3, and 4. You already know the answer should be 10, but let's look at two ways of thinking of this as a sequence and a partial sum for an arithmetic series.

**Method 1:**

If you think of the sequence as \( \{0, 1, 2, 3, \dots \} \) then \( a = 0 \) and \( d = 1 \). If you want to add the numbers up to 4 together, you would use \( n =5 \). That would give you the partial sum

\[ \begin{array}{lll} s_{5-1} &=& s_4 \\ &=& \sum\limits_{k=0}^{5-1} (0 + k\cdot 1) \\ &=& \sum\limits_{k=0}^{4} k \\ &=& 0 + 1 + 2 + 3 + 4 \\ &=& 10 , \end{array} \]

and if you used the formula

\[ s_{n-1} = s_4 = \frac{5}{2}(0 + (5-1)\cdot 1 ) = 10. \]

So this method gives the correct answer.

**Method 2: **

If you think of the sequence as \( \{1, 2, 3, \dots \} \) then \( a = 1 \) and \( d = 1 \). If you want to add the numbers up to 4 together, you would use \( n =4 \). That would give you the partial sum

\[ \begin{array}{lll} s_{4-1} &=& s_3 \\ &=& \sum\limits_{k=0}^{4-1} (1 + k\cdot 1) \\ &=& \sum\limits_{k=0}^{3} (1+k) \\ &=& (0 + 1) +( 1 + 1) + (1 + 2) + (1 + 3) \\ &=& 10 , \end{array} \]

and if you used the formula

\[ s_{n-1} = s_3 = \frac{4}{2}(1 + (4-1)\cdot 1 ) = 10. \]

So this method also gives the correct answer!

**What gives? ** Why do both methods give you the correct answer? That is because what is really happening is a substitution, just like in integration when you do a substitution and change the limits of integration.

**The key takeaway: ** When setting up a problem, it is always good to test out a shorter sum by hand to ensure you have the setup right before you go on to do the larger sum using the formula.

One of the most common ways to use the formula for the partial sums of an arithmetic series is to find the sum of the first \( n \) natural numbers.

Find the sum of the first 20 natural numbers.

Answer:

This question is really asking you to add the numbers 1, 2, 3, etc., all the way up to 20. Notice that each of those numbers is 1 unity apart, and from the deep dive above, you can think of this as the partial sum

\[ s_{21-1} = s_{20} = \sum\limits_{k=0}^{20} k , \mbox{ where } a=0, \quad d=1, \quad n=21,\]

or as

\[ s_{20 - 1} = s_{19} = \sum\limits_{k=0}^{19} (1+k) , \mbox{ where } a=1, \quad d=1, \quad n=20, \]

either one will give you the correct answer. If you choose the first one where \( a = 0 \), \( d = 1 \), and \( n = 21 \), then

\[ s_{20} = \sum\limits_{k=0}^{20} k = \frac{21}{2}\left(0 + (21 - 1)\cdot 1 \right) = 210. \]

So the sum of the first 20 natural numbers is 210.

Let's go back to the stadium example at the start.

You are building a football stadium, and each row has 10 fewer seats as you go toward the bottom of the stadium. Suppose that the stadium has 800 seats in the top row, and there are 25 rows total in the stadium. How many seats are there in the stadium?

Answer:

First, let's think of the first few rows and see what the pattern is.

Row 1: 800 seats

Row 2: 790 seats

Row 3: 780 seats

Now you do need to be a little careful because the formula for the arithmetic series starts with the 0^{th} row, not the first row. So instead, what you have is:

Row 0: 800 seats

Row 1: 790 seats

Row 2: 780 seats

This is an arithmetic sequence with \( a = 800 \) and \( d = -10 \).

You want to know how many seats there are in the stadium if there are 25 rows. What do you use for \( n\)?

Doing a little test, if there were only 4 rows of seats, you would add Row 0, Row 1, Row 2, and Row 3 to get a total of 4 rows of seats. That would make \( n = 4 \).

Since there are 25 rows in the stadium, you would use you can use the formula for the partial sums of an arithmetic series with \( n = 25 \) to get

\[ \begin{array}{lll} s_{25-1} &=& s_{24} \\ &=& \frac{25}{2} \left( 2\cdot 800 + (25 -1)\cdot (-10) \right) \\ &=& \frac{25}{2}(1600 - 240) \\ &=& 17000. \end{array} \]

So there will be 17000 seats in the stadium.

It can be easy to confuse geometric and arithmetic series. Just remember this:

a series is arithmetic if you

**subtract**consecutive terms and get a constanta series is geometric if you

**divide**consecutive terms and get a constant

When you think arithmetic series, think arithmetic. You learn to add and subtract first when doing arithmetic, so that is what goes with arithmetic series.

- An
**arithmetic sequence**has the form \( \{a, a+d, a+2d, a+3d, \dots \} \) where \(a \) and \( d \) are real numbers. - An
**arithmetic series is**one where the terms of the series are an arithmetic sequence. That means an arithmetic series looks like\[ \sum\limits_{n=0}^\infty (a + dn) . \]

- The partial sums for an arithmetic series have the form \[ s_{n-1} = \sum\limits_{k=0}^{n-1} (a+kd) = \frac{n}{2} (2a + (n-1)d) . \]
- The only arithmetic series that converges is the one that has terms that are all zero.

An arithmetic series is one where the difference between the terms of the sequence is constant.

It depends on the series, and how many terms of it you are summing up.

More about Arithmetic Series

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