### Topics

## Introduction to Series - At A Glance:

The arithmetic series is the one of the simplest series we can come up with. In terms of common restaurant menu items, the arithmetic series is a burger. Although it many be spiced up with bacon, feta cheese, and some type of questionable special sauce, it appears on every restaurant menu in one form or another. We need to understand this series type backward and forward.

An **arithmetic series** is a series whose terms form an arithmetic sequence. Simple enough, right? If we wrote down and arithmetic sequence, we could replace all of the commas with plus signs to get an arithmetic series.

### Sample Problem

The series

1 + 3 + 5 + 7 + ...

is an arithmetic series because

1, 3, 5, 7,...

is an arithmetic sequence.

### Sample Problem

The series

1 + 3 + 6 + 10 + ...

is not an arithmetic series because

1, 3, 6, 10, ...

is not an arithmetic sequence.

We have some good news. Just like a plain, old burger, we already know everything there is to know about the convergence of arithmetic series.

To start with, a constant series is an arithmetic series where the difference between successive terms is *d* = 0. The constant series

only converges if *a* = 0. That's a pretty bold statement. Take a look at a couple problems to understand why.

#### Example 1

Look at the arithmetic series Does this series converge or diverge? | |

Finding a formula for the partial sum *S*_{n} would be a bit annoying, but thankfully we don't need to bother with that. We know that *S*_{n} is the sum of the first *n* terms of the series: *S*_{n} = (4 + (1 – 1)3) + (4 + (2 – 1)3) + ... + (4 + (*n* – 1)3).
Each term is 4 plus something non-negative: *S*_{n} = (4 + (1 – 1)3) + (4 + (2 – 1)3) + ... + (4 + (*n* – 1)3).
Since each of the *n* terms we're adding up is at least 4, the partial sum is at least *n* times 4: *S*_{n} ≥ 4*n*.
From here it should be easy to see that as *n* approaches ∞, so does *S*_{n} (if this isn't easy to see, go review limits). Since is not a finite number, the sequence of partial sums diverges. This means the original arithmetic series diverges. | |

#### Exercise 1

Look at the arithmetic series

where

*a*_{i} = *a*_{1} + (*i* – 1)*d*

and both *a*_{1} and *d* are positive. Does this series converge or diverge?

Answer

This exercise is the same as the example except that we replaced some of the numbers with letters. The *n*th partial sum of the series is

*S*_{n} = (*a*_{1} + (1 – 1)*d*) + (*a*_{1} + (2 – 1)*d*) + ... + (*a*_{1} + (*n* – 1)*d*).

This is a sum of *n* terms. Since both *a*_{1} and *d* are positive, each term is a positive number greater than or equal to *a*_{1}. Thus

*S*_{n} ≥ *n* × *a*_{1}.

As *n* approaches ∞ so does *n* × *a*_{1}, which means

The sequence of partial sums diverges, which means the original arithmetic series diverges.

We made *a*_{1} positive in the exercise to make things easier to think about, but the series would still diverge if *a*_{1} were negative.

If *a*_{1} is negative and *d* is positive, eventually we'll add enough copies of *d* that we'll get a positive term

*a*_{n} = *a*_{1} + (*n* – 1)*d*.

From the exercise we know that, if our initial term and our difference *d* are both positive, the series diverges. Start the series out at *a*_{n} instead of *a*_{1}.

The series

has positive initial term *a*_{n} and positive difference *d*, so it diverges.

Since starting at a different term doesn't affect whether the series converges or diverges, the series

has to diverge also.

Our burger analogy makes more sense now. You can dress a burger with special sauce, lettuce, cheese pickles and onions, but it's still a burger. For our arithmetic burger series, it doesn't matter whether *a*_{1} is positive or negative. Either way, the series diverges.

#### Exercise 2

Look at the arithmetic series

where

*a*_{i} = *a*_{1} + (*i* – 1)*d*

and *d* is negative. Does this series converge or diverge?

Answer

If *a*_{1} and *d* are both negative, every term will be negative and at most *a*_{1}. In symbols,

*a*_{i} ≤ *a*_{1} < 0

for all *i*. If we sum *n* such terms, we see that the *n*th partial sum is

*S*_{n} ≤ *n*(*a*_{i}).

As *n* approaches infinity, the partial sums *S*_{n} keep getting more and more negative, so

This means the series diverges.

Just like the previous exercise, it doesn't matter if *a*_{1} is positive or negative. If we keep adding a negative value to *a*_{1}, eventually we'll get a negative term *a*_{n}. From what we just said, the series

must diverge. Since changing the starting limit of summation doesn't change whether the series diverges,

has to diverge also.

A constant series diverges unless the constant is 0. By comparing arithmetic series to a constant series, we showed that arithmetic series have to diverge unless both *a*_{1} and *d* are 0.