User:Dcljr/Series

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Introductory material from a complete rewrite of Calculus/Series

Infinite series[edit]

A series is the sum of a sequence of terms. An infinite series is the sum of an infinite number of terms (the actual sum of the series need not be infinite, as we will see below).

While there is usually some pattern in the terms being added, technically a series can be the sum of any sequence of terms:

1+2+\pi+\cos\tfrac{\pi}{12}+e+\ldots

The problem with the above series is, it is not at all clear what the next term of the series would be, nor any subsequent ones. To say anything useful about the series, there needs to be a clear pattern. Two such patterns seen in sequences that also appear in the study of series are called arithmetic and geometric.

An arithmetic series is the sum of a sequence of terms having a common difference (i.e., the difference between consecutive terms is always the same). For example,

1+4+7+10+13+\dots

is an arithmetic series with common difference 3, since a_2-a_1=3, a_3-a_2=3, and so forth. Unlike the first series above, this series has a clear pattern. It is obvious that the next term of the series is 16, and then 19, and so on.

A geometric series is the sum of a sequence of terms having a common ratio (i.e., the ratio between consecutive terms is always the same). For example,

1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots

is a geometric series with common ratio \tfrac{1}{2}, since a_2/a_1=\tfrac{1}{2}, a_3/a_2=\tfrac{1}{2}, and so forth. As before, there is a definite pattern, and the next term of the series is clearly \tfrac{1}{32}.

Summation notation[edit]

Summation notation provides a compact way of writing an infinite series. The arithmetic series above can be written

\sum_{n=1}^{\infty} [1+3(n-1)]

or, more simply

\sum_{n=1}^{\infty} (3n-2).

The symbol on the left is a Sigma (the Greek letter for capital "S", standing for "Sum"); it is also known as a summation symbol. Along with the expressions above and below the Sigma, it is read:

"The sum, as n goes from 1 to infinity, of..."

The algebraic expression on the right is the nth term of the series, and is often generically called "a_n", as is also done with sequences. This means that the sum could have been written a bit more verbosely (and perhaps more pedantically) as

\sum_{n=1}^{\infty} a_n \mbox{, where } a_n=3n-2,

a form that more explicitly reflects the idea that a series is the sum of a sequence of terms. (The expression a_n=3n-2 by itself defines the sequence of terms, but note that it is a bit ambiguous since it's not clear without additional context whether the sequence starts at n=1 or n=0, or indeed any other value. Incidentally, the first summation formula above, which might have seemed a bit strange at first, came from the formula for the nth term of an arithmetic sequence, which you probably first learned in algebra class: a_n=a_1+d(n-1).)

To verify that this summation formula represents the same sum as in the previous section, we simply evaluate the algebraic expression (a_n) for the first value of n (in this case, 1), then the next value (2), and so forth:

\sum_{n=1}^{\infty} (3n-2) = [3(1)-2]+[3(2)-2]+[3(3)-2]+\ldots = 1+4+7+\ldots

As for the geometric series, it should be easy to verify the representation

\sum_{n=1}^{\infty} \frac{1}{2^{n-1}}

or, equivalently

\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n-1}.

Non-uniqueness of summation notation[edit]

Note that the representation of a series in summation notation is not unique, even ignoring simple algebraic rewrites. All of the sums seen so far have started at n=1, as is usually the convention adopted for sequences, but we could have chosen to start the sums at any other value of n. In fact, it is quite common to start infinite series at n=0. If we follow that convention, the arithmetic series would be written

\sum_{n=0}^{\infty} (3n+1)

and the geometric series would be

\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n.

Again, these expressions should be easy to verify.

Furthermore, as with functions, the variable used in the sum (called the index variable) is completely arbitrary. Thus, the following sums also represent the same series (respectively) as those above:

\sum_{k=0}^{\infty} (3k+1)
\sum_{i=0}^{\infty} \left(\frac{1}{2}\right)^i

As a result of this fact, any series can be rewritten by simply changing the starting value of the series and compensating for this change in the formula for the nth term. To be specific, consider the first series above in this section. Using the substitution n=m-2, we get the series

\sum_{m-2=0}^{\infty} (3[m-2]+1) = \sum_{m=2}^\infty (3m-5).

But then the variable m can simply be changed back to n to get

\sum_{n=2}^{\infty} (3n-5).

Notice how this new series can be seen as resulting from "bumping up" the starting value of the original series by 2 and "bumping down" each n in the a_n formula by the same amount (that is, replacing each n by n-2 and then simplifying). This kind of change to a series is sometimes necessary to verify a given series formula (say, when looking up the answer to an exercise in a textbook).

Series versus sequence[edit]

As mentioned above, there is an important difference between an infinite series, which is the sum of an infinite number of terms, and the sequence formed by those same terms.

At the risk of being repetitive, our arithmetic series is

\sum_{n=1}^{\infty} (3n-2) = 1+4+7+10+\ldots,

but the corresponding sequence of terms is

\{a_n\}_{n=1}^{\infty} = \{3n-2\}_{n=1}^{\infty} = 1,4,7,10,\ldots.

It might seem silly at this point to make such a big deal out of this distinction, but it will be very important going forward not to confuse these two concepts.

Sequence of partial sums[edit]

One special kind of sequence that is very useful to consider when studying series is the sequence of partial sums, defined in the following way (assuming a starting value of 1 for the index variable) for any positive integer n:

s_n=\sum_{k=1}^{n}a_k=a_1+a_2+a_3+\ldots+a_n

For our arithmetic series, the sequence of partial sums is:

s_1 =a_1=1
s_2 =a_1+a_2=1+4=5
s_3 =a_1+a_2+a_3=1+4+7=12
s_4 =a_1+a_2+a_3+a_4=1+4+7+10=22
\ldots

Thus

\left\{s_n\right\}=1,5,12,22,\ldots.

Note how this sequence of partial sums is very different from the sequence of terms discussed in the last section!

Just as the series itself can be written more compactly in summation notation by finding the pattern in its terms, the sequence of partial sums for a series can (sometimes) also be written compactly as a function of n.

It should be easy to check that the following expression accurately represents the (first few terms of the) sequence of partial sums shown above:

s_n =\frac{3n^2-n}{2}

To see where this expression came from, we first need to review some properties of sums that you probably remember from arithmetic and algebra, but might not be familiar with in summation notation.

Some properties of finite sums[edit]

Constant factors can be factored out of (or multiplied into) finite sums:

\sum_{k=1}^{n} c\,a_k=c\sum_{k=1}^{n}a_k

Finite sums can be added and subtracted, as long as they cover the same range of values of the index varible:

\sum_{k=1}^{n} a_k+\sum_{k=1}^{n}b_k=\sum_{k=1}^{n}(a_k+b_k)
\sum_{k=1}^{n} a_k-\sum_{k=1}^{n}b_k=\sum_{k=1}^{n}(a_k-b_k)

Putting these two ideas together, one may show the following property for arbitrary linear combinations:

\sum_{k=1}^{n} (c\,a_k+d\,b_k)=c\sum_{k=1}^{n}a_k+d\sum_{k=1}^{n}b_k

Furthermore, there are special formulas for sums of certain simple expressions:

\sum_{k=1}^{n} 1=n
\sum_{k=1}^{n} k=\frac{n(n+1)}{2}
\sum_{k=1}^{n} r^k=\frac{r(1-r^n)}{1-r}\mbox{ if }r\ne1

The first formula is obvious, since it represents the sum of n copies of the number 1. You might recognize the second formula as arising in the study of the binomial theorem and combinations (in particular, it represents the number of possible combinations of n distinct objects taken 2 at a time). The third formula is usually encountered in intermediate algebra or precalculus classes when geometric series are first studied.

With these facts, one may derive the expression given above for the sequence of partial sums of our arithmetic sequence.

\begin{align}
\sum_{k=1}^{n} (3k-2)&=\left(\sum_{k=1}^{n} 3k\right) - \left(\sum_{k=1}^{n} 2\right) \\
 &=3\left(\sum_{k=1}^{n} k\right) - 2\left(\sum_{k=1}^{n} 1\right) \\
 &=3\left[\frac{n(n+1)}{2}\right]-2(n) \\
 &=\frac{3n^2+3n}{2}-\frac{4n}{2} \\
s_n&=\frac{3n^2-n}{2}
\end{align}

Similarly, one may derive the sequence of partial sums for the geometric series:

\begin{align}
\sum_{k=1}^{n} (\tfrac{1}{2})^{k-1} &= \sum_{k=1}^{n} (\tfrac{1}{2})^k(\tfrac{1}{2})^{-1} \\
 &= \sum_{k=1}^{n} (\tfrac{1}{2})^k(2) \\
 &= 2\sum_{k=1}^{n} (\tfrac{1}{2})^{k} \\
 &= 2\left(\frac{\tfrac{1}{2}[1-(\tfrac{1}{2})^n]}{1-\tfrac{1}{2}}\right) \\
s_n &= 2[1-(\tfrac{1}{2})^n]
\end{align}

Finding a series from its partial sums[edit]

Using the fact that

s_n=a_1+a_2+\ldots+a_{n-1}+a_n=s_{n-1}+a_n,

we see that

a_n=s_n-s_{n-1}.

This provides a way of "recovering" the original series from its sequence of partial sums.

For our arithmetic series:

\begin{align}
a_n &= s_n-s_{n-1} \\
 &=\left[\frac{3n^2-n}{2}\right]-\left[\frac{3(n-1)^2-(n-1)}{2}\right] \\
 &=\left[\frac{3n^2-n}{2}\right]-\left[\frac{3(n^2-2n+1)-(n-1)}{2}\right] \\
 &=\left[\frac{3n^2-n}{2}\right]-\left[\frac{3n^2-6n+3-n+1}{2}\right] \\
 &=\frac{3n^2-n-3n^2+6n-3+n-1}{2} \\
 &=\frac{6n-4}{2} \\
a_n&=3n-2
\end{align}

Sum of an infinite series[edit]

Now we can finally formally define the sum of an infinite series as the limit of its sequence of partial sums:

\sum_{n=1}^{\infty} a_n=\lim_{n\to\infty}s_n

If the limit converges to a real number, say s, then the infinite series is said to converge to the sum s, or to be convergent with sum s. If the limit diverges (including the cases where the limit is infinity or negative infinity), then the series is said to diverge or to be divergent in the same way, and its sum is said to not exist (or to be infinity or negative infinity, as appropriate). Note that one does not describe the series itself as "not existing" when it diverges, only its sum.

Consider again the arithmetic and geometric series we have been discussing up to this point. It is obvious by simply looking at the original arithmetic series

1+4+7+10+13+\dots

that it does not have a finite sum. The terms being added are themselves getting larger and larger without bound, so the sum is getting larger and larger without bound.

The sequence of partial sums given above formalizes this idea. In particular, because

\sum_{n=1}^{\infty} (3n-2) = \lim_{n\to\infty} \frac{3n^2-n}{2} = \infty,

the arithmetic series diverges to infinity.

Now consider the original geometric series:

1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots

The terms here are getting smaller and smaller, and indeed are approaching zero. Since adding zero to something doesn't change its value, it seems reasonable to suspect that the sum might be a fixed, finite number. What does the sequence of partial sums reveal?

\sum_{n=1}^{\infty} (\tfrac{1}{2})^{n-1} = \lim_{n\to\infty} 2[1-(\tfrac{1}{2})^n] = 2(1-0)=2

So the geometric series does, in fact, converge to the finite sum 2.

It seems obvious that any series whose terms "blow up" to infinity (like our arithmetic series) will diverge, but does every series whose terms shrink to zero (like our geometric series) converge to a finite sum? It turns out the answer to that question is no.

Harmonic series[edit]

There is a very important series whose sequence of terms goes to zero and yet the series diverges because the sequence of partial sums diverges to infinity. It is called the harmonic series:

\sum_{n=1}^{\infty} \frac{1}{n}

Obviously the terms of this series go to zero:

\lim_{n\to\infty} \frac{1}{n}=0

But what about the sequence of partial sums? For convenience sake, we consider not the nth partial sum but the 2^nth partial sum, then group the terms in a clever way, and find a lower bound for each group of terms:

\begin{matrix}
\displaystyle{\sum_{k=1}^{2^n} \frac{1}{k}} &= 1+{} &\underbrace{\frac{1}{2}}_{\text{group }1} &+ &\underbrace{\frac{1}{3}+\frac{1}{4}}_{\text{group }2} &+ &\underbrace{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}}_{\text{group }3} &{}+\ldots+{} &\underbrace{\sum_{k=2^{n-1}+1}^{2^n} \frac{1}{k}}_{\text{group }n} \\
 &> 1+{} &\frac{1}{2} &+ &\frac{1}{4}(2) &+ &\frac{1}{8}(4) &{}+\ldots+{} &\frac{1}{2^n}(2^{n-1}) \\
\\
 &= 1+{} &\frac{1}{2} &+ &\frac{1}{2} &+ &\frac{1}{2} &{}+\ldots+{} &\frac{1}{2}
\end{matrix}

There are n terms equal to \tfrac{1}{2} in the final sum, thus

\sum_{k=1}^{2^n} \frac{1}{k} > 1+\sum_{k=1}^{n} \frac{1}{2}=1+\frac{n}{2}.

The limit of this final expression, as n\to\infty, is infinity. This means the sequence of partial sums diverges. (Technically, we have only shown that a "subsequence" of the original sequence diverges, but it turns out that this is sufficient to prove that the original sequence diverges.)

Therefore, the harmonic series diverges even though its terms shrink to zero. Many other series share this property, some of which will be discussed below.

"Eventually"[edit]

When considering whether a given series is convergent or divergent, it is often useful to "ignore" the first several terms of the series. For example, as we have seen, the series

1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots

is geometric and converges to 2, but the series

-1+3+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots

is not geometric, since the first two terms don't fit the pattern of the rest of the series. But since the series behaves like a geometric series from the third term onward, one might call the series "eventually" geometric: it didn't start out that way, but eventually it settled down and behaved like a geometric series. We will use this idea of a series "eventually" having a certain property many times in the discussion that follows.

Clearly if the first (geometric) series above converges, the second (eventually geometric) series will also converge. In fact, by considering separately the sum of the first two terms and the sum of the rest of the terms, we can deduce that the sum of the second series is

-1+3+\sum_{n=1}^{\infty} \frac{1}{2^{n-1}} = (-1+3)+(2) = 4.

Similarly, the series

\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots

converges in the same way as the original geometric series, but again the new (this time shorter) series converges to a different sum (the original sum minus the first two terms):

\sum_{n=3}^{\infty} \frac{1}{2^{n-1}} = \biggl(\sum_{n=1}^{\infty} \frac{1}{2^{n-1}}\biggr) - \biggl(\sum_{n=1}^2 \frac{1}{2^{n-1}}\biggr) = (2)-(1+\tfrac{1}{2}) = \tfrac{1}{2}

So, note that whether a series converges doesn't depend on what's happening at the beginning of the series, but the sum of the series definitely does!


Material about convergence and divergence tests still being written…