What is a series

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adding a sequence

In the previous sections, we’ve seen several examples of sequences. If we have a sequence $\{a_n\}_{n=1}$, then we can ask two important questions about it.

Do the numbers in the list approach a finite value?
If so, what is the value?

The first question is called an existence question. Does the limit exist? This is a yes or no question.

If the answer is yes (the limit exists), then we come to the second question.

In this section, we propose a different view of a sequence of numbers. We can view them as terms in a partial summation, where we keep adding on more terms, like digits in a decimal expansion.

Consider the sequence $(a_n)_{n=1} = 5, 8, 7, 9, 10, 10, 13, \cdots $.

This sequence could be viewed as

\begin{align*} a_1 & = 5 & = 5 \\ a_2 & = 8 & = 5 + 3 \\ a_3 & = 7 & = 5 + 3 - 1 \\ a_4 & = 9 & = 5 + 3 - 1 + 2 \\ a_5 & = 10 & = 5 + 3 - 1 + 2 + 1 \\ a_6 & = 10 & = 5 + 3 - 1 + 2 + 1 + 0 \\ a_7 & = 13 & = 5 + 3 - 1 + 2 + 1 + 0 + 3 \end{align*}

The terms of the original sequence, $(a_n)$, are each the sum of the first $n$ numbers from the sequence

\[ (b_k)_{k=1} = 5, 3, -1, 2, 1, 0, 3, \cdots \]

We could go the other way as well. We could begin with a sequence and then create another sequence where the terms are the partial sums of the original sequence.

Consider the sequence $(a_k)_{k=1} = 3, 6, 2, 1, 0, 4, \cdots $.

Define a new sequence $(s_n)_{n=1}$ as

\[ (s_n)_{n=1} = a_1 + a_2 + a_3 + \cdots + a_{n-1} + a_n \]

\begin{align*} s_1 & = 3 \\ s_2 & = 3 + 6 & = 9 \\ s_3 & = 3 + 6 + 2 & = 11 \\ s_4 & = 3 + 6 + 2 + 1 & = 12 \\ s_5 & = 3 + 6 + 2 + 1 + 0 & = 12 \\ s_6 & = 3 + 6 + 2 + 1 + 0 + 4 & = 16 \\ s_7 & = \end{align*}

The sequence $(s_n)_{n=1}$ is a a sequence of partial sums of $(a_k)_{k=1}$.

We can ask an important question about a sequence of partial sums.

Can we sum all of the numbers in the first list and obtain a finite result?

Another existence question.

Does limit $\lim \limits _{n \to \infty } s_n$ exists ?

On one hand, this is a question about the convergence or the sequence $(s_n)_{n=1}$. On the other hand, this is a weird question about adding up an infinite number of terms from the original sequence.

Let $(a_k)_{k=1}$ be a sequence.

Let $(s_n)_{n=1} = a_1 + a_2 + a_3 + \cdots + a_{n-1} + a_n$ be the sequence of partial sums.

We have shorthand notation for the partial sums.

\[ s_n = \sum _{k=1}^n a_k = a_1 + a_2 + a_3 + \cdots + a_{n-1} + a_n \]

Shading A Square

Suppose that we want to study the infinite sum below.

\[ \frac {1}{2} + \left (\frac {1}{2}\right )^2+ \left (\frac {1}{2}\right )^3+ \cdots \]

A student, feeling quite clever, decides to illustrate the sum by drawing a square with side length one unit and shading it in a special way.

Step 1: Shade the left half of the square.

There are two quantities of which we can keep track now.

Call the area shaded in this step $A_1$. Then, we have $A_1=\answer [given]{1/2}$ square units.

Step 2: Shade the bottom half of the unshaded region.

Call the new area shaded in this step $A_2$. Then, we have $A_2=\answer [given]{1/4}$.
Call the total shaded area $S_2 = A_1 + A_2$. Then, we have $S_2=\answer [given]{3/4}$.

Visually, notice that we can find $S_2$ by noting

\[( \textrm {area shaded after Step } 2) = ( \textrm {area from Step } 1) + (\textrm {area from Step } 2)\]

and writing down the total shaded are of the square.

Analytically, we can write $S_2=A_1+A_2.$

Step 3: Shade the left half of the unshaded region.

Call the new area shaded in this step $A_3$. Then, we have $A_3=\answer [given]{1/8}$.
Call the total shaded area $S_3$. Then, we have $S_3=\answer [given]{7/8}$.

We can think of $S_3$ visually or analytically.

\[ S_3 = A_1 + A_2 + A_3 \]

Hopefully, the pattern used to shade the square is becoming clear. We can define the new area shaded during the $n$-th step to be $A_n$, and can observe that $A_n = \left ( \frac {1}{2} \right )^n$.

We can also let $S_n$ denote the total shaded area after the $n$-th step. Analytically, we have $S_n = A_1+A_2 + \ldots A_n$, or by using summation notation, we can write $\displaystyle S_n = \sum _{k=1}^n A_k$.

Looking at the pictures drawn so far, notice that the only unshaded area after the $n$-th step is a rectangle of area $A_n$, so we can write a formula for $S_n$.

\[ S_n = \left <\textrm {area of the whole square}\right >-A_n = 1-\left (\frac {1}{2}\right )^n \]

We can now evaluate the limit and find that $\lim \limits _{n \to \infty } S_n =\answer [given]{1}$.

We also have another method of thinking about this limit; after we continue shading the square indefinitely, there will be no portion of it that has not been shaded. Thus, the total shaded area should be $1$.

We would thus like to conclude that

\[ \frac {1}{2} + \left (\frac {1}{2}\right )^2+ \left (\frac {1}{2}\right )^3+ \ldots = \sum \limits _{k=1}^{\infty } \left (\frac {1}{2}\right )^n =1. \]

Infinite series

The above scenario can be modeled using sequences. We have $\{A_n\}_{n=1}$ whose $n$-th term is given by the explicit formula $A_n=\left (\frac {1}{2}\right )^n$, and we represent the sequence by the ordered list below.

\[ \frac {1}{2},\left (\frac {1}{2}\right )^2,\left (\frac {1}{2}\right )^3,\ldots \]

and we can interpret the series

\[\frac {1}{2} + \left (\frac {1}{2}\right )^2+ \left (\frac {1}{2}\right )^3+ \ldots \]

as the attempt to add up all of the terms in the above sequence. In order to determine if this can be done, we found a new sequence $\{S_n\}_{n=1}$ that was created from the original sequence. It is represented by the list below.

\[ \frac {1}{2},\frac {3}{4},\frac {7}{8},\ldots \]

Note that the limit of this new sequence is exactly the sum of all of the terms in the old sequence! Let’s formalize the ideas in the last example with a definition.

Sequence of Partial Sums

Let $\{a_n\}_{n=n_0}$ be a sequence. Let $s_n = \sum \limits _{k=k_0}^n a_k$; the sequence $\{s_n\}_{n=k_0}$ is the called the sequence of partial sums of $\{a_n\}$.

(a): The series $\sum \limits _{k=k_0}^\infty a_k$ converges if and only if $\lim \limits _{n\to \infty } s_n$ exists. Furthermore, if $\lim \limits _{n\to \infty } s_n =L$, we say the series $\sum \limits _{k=k_0}^\infty a_k$ converges to $L$.
(b): The series $\sum \limits _{k=k_0}^\infty a_k$ diverges if and only if $\lim \limits _{n\to \infty } s_n = \infty , \lim \limits _{n\to \infty } s_n = -\infty $ or $\lim \limits _{n\to \infty } s_n $ otherwise does not exist.

The above definition really states that the symbols $\sum \limits _{k=k_0}^{\infty } a_k$ and $\lim \limits _{n \to \infty } s_n$ represent exactly the samething! It makes the content of the previous example more precise. This is quite important since we have techniques to determine whether limits of sequences exist. We are now able to recast the new question “Can I sum all of the terms in a sequence?” into the old question “Does the associated sequence of partial sums have a limit?”. To attack the latter, we can utilize all of our previous techniques for studying the limit of the sequence of partial sums.

Some people define the $n$-th term in the sequence of partial sums by summing the first $n$ terms in the original sequence. With our convention, the lower index for both the original sequence and the sequence of partial sums will always be the same; that is, both sequences share a common domain. In the case when the lower index is $k_0=1$, both conventions are equivalent.

Using our new terminology, what can we say about the series $\sum \limits _{k=1}^\infty \left (\frac {1}{2}\right )^k$ from the previous example? The series convergesdiverges, and $\sum \limits _{n=1}^\infty \left (\frac {1}{2}\right )^n = \answer [given]{1}$.

Now, let’s see an example.

Suppose that $\{a_n\}_{n=1}$ is a sequence and let $s_n = \sum \limits _{k=1}^n a_k$. Suppose it is known that

\[ s_n = \frac {\ln (n)+5n^2}{2n^2+1}. \]

Determine whether $\sum \limits _{k=1}^{\infty } a_k$ converges or diverges. If it converges, give its value.

The symbols $\sum \limits _{k=1}^{\infty } a_k$ and $\lim \limits _{n \to \infty } s_n$ are equivalent; in order to determine whether $\sum \limits _{k=1}^{\infty } a_k$ converges or diverges, we need to analyze $\lim \limits _{n \to \infty } s_n$.

The dominant term in the numerator is $\ln (n)$$5n^2$.
The dominant term in the denominator is $2n^2$$1$.

We can conclude that

\[ \lim \limits _{n \to \infty } s_n = \lim \limits _{n \to \infty } \frac {\ln (n)+5n^2}{2n^2+1} = \lim \limits _{n \to \infty } \frac {5n^2}{2n^2} = \answer [given]{\frac {5}{2}}\]

Hence, $\sum \limits _{k=1}^{\infty } a_k$ converges since $\lim \limits _{n \to \infty } s_n$ exists, and since $\lim \limits _{n \to \infty } s_n=\frac {5}{2}$, we have that $\sum \limits _{k=1}^{\infty } a_k$ converges to $\frac {5}{2}$.

Properties of sums

We finish this section by giving some properties of series.

Sums and Constant Multiples of Convergent Series Let $\sum \limits _{k=k_0}^\infty a_k$ and $\sum \limits _{k=k_0}^\infty b_k$ be convergent series and suppose that $ \sum \limits _{k=k_0}^\infty a_k = A$ and $\sum \limits _{k=k_0}^\infty b_k =B.$

(a): Constant Multiple Rule
For any constant $c$,
\[\sum \limits _{k=k_0}^\infty c\cdot a_k = c\cdot \sum \limits _{k=k_0}^\infty a_k = c\cdot A.\]
(b): Sum/Difference Rule
\[\sum \limits _{k=k_0}^\infty \big (a_k\pm b_k\big ) = \sum \limits _{k=k_0}^\infty a_k \pm \sum \limits _{k=k_0}^\infty b_k = A \pm B.\]

Notice, of course, that we’re working with convergent series in this theorem. Adding divergent series is trickier, but there is something we can say about the attempt to add a convergent series and a divergent series.

Suppose $\sum \limits _{k=k_0}^\infty a_k$ converges and $\sum \limits _{k=k_0}^\infty b_k$ diverges. Then, $\sum \limits _{k=k_0}^{\infty } \big (a_k+b_k\big )$ diverges.

To understand why this theorem holds, note that if $\sum \limits _{k=k_0}^\infty \big (a_k+b_k\big )$ would converge, by the last theorem, we would know that the difference

\[\sum \limits _{k=k_0}^\infty \big (a_k+b_k\big ) -\sum \limits _{k=k_0}^\infty a_k\]

converges since both series above are convergent. Furthermore, the previous theorem also guarantees that

\[\sum \limits _{k=k_0}^\infty \big (a_k+b_k\big ) -\sum \limits _{k=k_0}^\infty a_k = \sum \limits _{k=k_0}^\infty \big (a_k+b_k -a_k\big ),\]

but this is precisely $\sum \limits _{k=k_0}^{\infty } b_k$, which diverges by assumption. This is a contradiction, so $\sum \limits _{k=k_0}^\infty \big (a_k+b_k\big )$ must diverge.

Convergence of Tails Let $\{a_n\}_{n=n_0}$ be a sequence and $n_1 \geq n_0$ be an integer. Then, $\sum \limits _{k=n_0}^{\infty } a_k$ and $\sum \limits _{k=n_1}^{\infty } a_k$ either both converge or both diverge.

Essentially, this theorem ensures that it does not matter where we start summing the terms of a sequence. Since we have infinitely many terms to try to add together, the sum of the first finitely many will not affect if this addition is possible.

If $n_1>n_0$, the series $\sum \limits _{k=n_1}^{\infty } a_k$ is often called a tail of the series $\sum \limits _{k=n_0}^{\infty } a_k$.

We now finish the section with an example that ties many ideas together.

Suppose that $\{a_n\}_{n=1}$ is a sequence and define its sequence of partial sums $\{s_n\}_{n=1}$ by the usual rule $s_n = \sum \limits _{k=1}^n a_k$. Suppose it is known that

\[ s_n = \frac {8n^2}{n^4-9}. \]

What is $a_1+a_2+a_3$? $1$$\frac {18}{7}$It is not defined.

Note that we are given an explicit formula for the term $s_n$. By definition, we have $s_3 = a_1+a_2+a_3$, so all we have to do to find $a_1+a_2+a_3$ is to evaluate the formula for $s_n$ when $n=3$.

\[ a_1+a_2+a_3 = s_3 = \frac {8(3)^2}{(3)^4-9} = 1 \]

What is $a_2+a_3$?

Note that $s_3 = a_1+a_2+a_3$ and $s_1 = a_1$. Thus,

\[s_3 -s_1 = \left (\cancel {a_1}+ a_2+a_3\right ) -\cancel {a_1} = a_2+a_3\]

Using the formula for $s_n$ gives $s_3 = \answer [given]{1}$ and $s_1=\answer [given]{-1}$, so $a_2+a_3 = \answer [given]{2}$.

Determine whether $\sum \limits _{k=1}^{\infty } a_k$ converges or diverges. If it converges, what is its value?

We can determine whether $\sum \limits _{k=1}^{\infty } a_k$ converges or diverges by analyzing $\lim \limits _{n \to \infty } s_n = \lim \limits _{n \to \infty } \frac {8n^2}{n^4-9}$. Since this limit is zero, we know that $\sum \limits _{k=1}^{\infty } a_k$ converges to $0$.

Determine whether $\sum \limits _{k=4}^{\infty } a_k$ converges or diverges. If it converges, can you find its value?

Since $\sum \limits _{k=1}^{\infty } a_k$ converges, changing the lower index to $k=4$ willwill not affect whether the series converges. Since we found that $\sum \limits _{k=1}^{\infty } a_k$ converges, so too will $\sum \limits _{k=4}^{\infty } a_k$. However, since $k=4$ is our lower index here, it willwill not potentially affect the value of the series.

Let’s write out the series and make a few observations.

Putting this together, we have:

\[ 0 = s_3 + \sum \limits _{k=4}^{\infty } a_k, \]

and we find that $ \sum \limits _{k=4}^{\infty } a_k = -1$.

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more examples can be found by following this link
More Examples of Sums of Sequences

2025-05-18 03:52:38