© 2000 M.
Flashman

We begin our next discussion with a sequence that has its terms determined as sums. Its properties illustrate much of what is typical for nicely behaved functions.

**Example X.B.1.** Let

`S_n = sum_{k=0}^{k=n} (3/4)^k`

Multiplying the first equation by `3/4` gives

`3/4S_n = 3/4+ (3/4)^2 +(3/4)^3 + ... +(3/4)^n + (3/4)^{n+1}`.

Taking the difference of these two expressions we obtain

`1/4 S_n = 1 - (3/4)^{n+1}`,

so we see finally that `S_n = 4 - 4 (3/4)^{n+1}`. From
our previous work on geometric
sequences, the limit of `S_n` is
`4`, *i.e*., `lim_{ n->oo}S_n
= 4`.

We generalize this last example in the following theorem:

**Theorem X.B.1** **(Convergence of geometric series.)**

Suppose `S_n = a + ar+ ar^2 + ... +ar^n = a(sum_{k=0}^{ k=n} r^k)`

. Then

`lim_{ n->oo}S_n =a/{1-r}` if and only if `|r|<1`.

**Proof**: Following the algebra used in the example we multiply
S_{ n} by *r*

`- rS_n =` ` ar+ ar^2 +... +ar^n+ar^{n+1}`

`(1 - r) S_n = a - ar^{n+1}`

As in the example then,

the sequence `S_n` will converge if
and only if the geometric sequence `r^{n+1}` converges (with
`r`
not equal to `1`), i.e., if and only if `-1 < r < 1`.

**Comments:**

- To see a dynamic java sketch that visualizes this result for `0< r < 1`, click here.
- Notice that by replacing `r` with `x` we have that `S_n` is a
polynomial
, in fact the MacLaurin polynomial for the function
`f
(x)
= 1/(1-x)` . That is,

`S_n = P_n(x,f )`. - The difference between `S_n` and its limit for any particular `r` is given precisely by the expression `a{r^{n+1}}/{1 - r}`. When `-1<r<1` this difference gives the precise error in using `S_n` to estimate `a/{1-r}`.
- To generalize this example we can consider (1) sequences of sums of numbers or more specifically (2) sequences of sums of numbers that arise as multiples of powers of a specific number.
- The result in this situation said that when `S_n` is a polynomial function in the variable `x`, the sequence of sums would converge for values of `x` in a specific interval and would diverge for all other `x`. This situation will generalize as well. For each sequence of polynomials there will be an interval of values for the variable on which the sequence will converge and outside of which the sequence will diverge.

**Example X.B.2.** Find** the limit of the geometric
series `**3 + 1.2 + .48 + ...`** , i.e., find **`lim_{n->oo}(3
+ 1.2 + .48 + ... )`.

**Solution: **The main issue here is to recognize this problem
as
an application of the **geometric series** result just
proven.
The first term in the sequence of a geometric series is `a` and the
ratio
of any consecutive summands is the `r`. Thus we have that for this
example
`a = 3` and `r = 1.2/3 = .48/1.2 = .4` . The limit is

**B.2 The (Positive) Harmonic Series.**

Here is another important example. It is the sequence that arises from considering sums of the reciprocals of the natural numbers. Although the individual summands in this example are approaching zero, the sequence of sums is increasing without bound, so it cannot have a limit.

**Example X.B.3.** **The Harmonic Series. **If `S_n
= 1 + 1/2 + 1/3 + ... + 1/n = sum_{k=1}^{k=n} 1/k` for `n = 1, 2, 3,
...` then the sequence
`S_n` diverges. This can be understood most easily by recognizing
that `S_{n+1} > S_n` and that when `n = 2^k`
we have

`S_n = 1 + (1/2) + (1/3+1/4) + (1/5 + ... + 1/8) + ... + 1/n`

`>1 + (1/2)+ (1/4+1/4) + (1/8 + ... + 1/8) + ... + 1/{2^k}``>1 + (1/2) + (2/4) + (4/8) + ... + (2^{k-1}/{2^k})`

`>1 + k 1/2`.

Thus by taking a large enough value for `n = 2^k`, the value
of `S_n` can be made larger than any specified number. So the
sequence `S_n` diverges.

**Comments**: 1. Another way to understand the divergence of the
geometric sequence is to notice that for each `k` the summand term
`1/k` can
be considered as the area of the region in the plane contained above
the
`X`-axis and below the line `Y = 1/k` between the lines `X = k` and `X
= k+1`.

Since the function `f (x) = 1/x` is decreasing on this interval, the region just described contains the region in the plane contained above the `X`-axis and below the curve `Y = 1/X` between the lines `X=k` and `X = k+1`. Thus

`1/k > int_{x=k}^{x=k+1}1/x dx` .

Adding each of these inqualities to consider `S_n` gives

`S_n >sum_{k=1}^{k=n} int _{x=k}^{x=k+1}1/x dx =int _{x=1}^{x=k+1} 1/x dx =ln(n+1)` .

But we know that by choosing n large we can make the expression `ln
(n+1)`
as large as desired,

so `S_n` diverges.

2. The techniques used to investigate the harmonic series can be generalized in several different ways to investigate other series with positive summands. What was crucial was the ability to compare the individual summands in the series to numbers whose sum could be recognized to grow large without bound.

3. It is very important to remember the harmonic series to avoid making the common mistake of believing that a sequence of sums has a limit merely because the individual summands in the sequence are approaching zero .

It is remarkable that a slight variation of the sequence of sums of
reciprocals will converge.

The introduction of a changing (alternating)
sign with the factor `(-1)^ {n+1}` is the key to this kind of
situation.

**Example X.B.4.** **The Alternating Harmonic Series**.

Let `S_n = 1 - 1/2 + 1/3 - ... + (-1)^{n+1}1/n` for `n = 1, 2, 3, ...` then `S_n` {`n = 1, 2, ...` } converges.

**Discussion.** First we notice that `S_1 = 1 > S_3
> S_5 > ... > S_ n > 1/2` where `n` is an odd number.
[Can you explain this?] Thus the sequence of values of

`S_n` when
`n` is an odd number is a decreasing sequence of numbers that is
bounded
below by the number `1/2`. Therefore this *sub*sequence of the
original
sequence has a limit, which we will call `bar S`.

Suppose `m` is an even number, say
`2k`, then `m` is `1` more than an odd number,that is, there is an odd
number `n` where`n+1 = 2k = m`. Now in this case

`S_m = S_{2k} = S_{n+1} = S_n
+ {(-1)^{ n+2}} /{n+1}` and thus when `n -> oo` ,

**Comments:** 1. The technique used here can be generalized to
explain
the convergence of many other alternating sequences. What was needed
for
the argument was that the individual summands alternated in sign while
the magnitude of the individual summands, `1/{n+1}`, was decreasing to
0. The argument also relied
on the fact that a decreasing sequence of numbers that is bounded below
must converge to some number.

2. The alternating harmonic sequence is related to the Taylor polynomial for the natural logarithm function about `x = 1`. This polynomial is

When `n` is large and `1 < x < 2` the remainder term for the Taylor polynomial can be shown to approach `0`, so when `1 < x < 2` we can say that

`lim_{n->oo}P_n(x,ln(x),1) = ln(x)`.

Now we can use the continuity of polynomial functions and the natural logarithm to complete the argument: When `x` is close to `2`, `P_n(x,ln(x), 1)` is close to `P_n(2,ln(x),1)` and `ln(x)` is close to `ln(2)` while when `n` is large, `P_n(x,ln(x),1)`, is close to `ln(x)`.

But the sequence `P_n(2,ln(x), 1), n = 0, 1, 2, ...` is precisely the alternating harmonic series. So when `n` is large and `x` is close to `2`, the sums `P_n(x,ln(x), 1)` are close to `ln(2)` while also being close to the value `bar S` discussed as the limit of the alternating harmonic series. But there can be only one limit for a sequence, so we have that `bar S = ln (2)`.

**B.4 Terminology and Notation for Infinite Series.**

**Infinite Series:** In the last sections we have examined
sequences
that arose from sums of numbers. These sequences of sums have
traditionally
been called infinite series. If we denote the individual summands with
subscripts we might say that

`S_n = c_0 + c_1 + c_2 + ...
+ c_ k + ... + c_n = sum _{k=0}^{k=n} c_k_{ `} where the
numbers `c_k` are the summands.

For a particular `n`, `S_n`**
is called the partial sum of the
series**.

If the sequence `S_n` converges to a limit, `S`, we
say the series converges to `S`.

If the series does not converge, we say
the series diverges.

The series in its entirety is denoted formally by
the expression `sum_{k=0}{oo}c_k`.

**Warning**: **Be careful not to presume that when we express
an
infinite series with this consolidated notation the series is convergent**.

When the series converges to a limit, `S`, we write

`sum_{k=0}^{oo}c_k = S`.

**Example X.B.5.** To summarize the work in the previous section
with this notation, we have the following results:

**Geometric Series :**

When `| x | < 1` then `sum_{k=0}^{oo}ax^k = a/{1-x}`.

**Harmonic Series:**

`sum_{k=1}^{oo}(-1)^{k+1}1/k = ln (2)`.

**Power Series. **The geometric series is typical of a type of
infinite
series that is particularly important to our study of the calculus. Its
individual summands are each nonnegative integer powers of the variable
`x`. Its partial summands are polynomial functions in the variable `x`.
This
is precisely the situation we have encountered in studying MacLaurin
and
Taylor polynomials.

We call an infinite series - a power series in `x` if for each `n` the partial summand `S_n` is a polynomial function of degree `n`.

We say that** **a power series
in `x` converges in (or on)
an interval
I if the infinite series converges for any choice of `x` in the
interval
I.

One of our chief purposes in introducing the topics of sequences and
series is to understand more fully the meaning of the Taylor theory.

For
example, given a `C^{oo}`
function `f` on an interval I, containing `0`, we can consider
immediately the
power series determined by the Maclaurin polynomials of that function.
This power series is descibed as the Maclaurin (or Taylor) series for
`f`
(at `x = a`) and is expressed simply as

Important questions for the Taylor theory concern this power series. For example, (1) for which values of `x` will the series converge? and (2) when the series does converge, does `P(x,f ) = f (x)` ? The next example summarizes some results about many important functions and their related Maclaurin series.

**Example X.B.6.** (Important Taylor Series Converge.)

`cos(x) =sum _{k=0}^{oo}(-1)^k {x^{2k}}/{(2k)!}`.

`sin(x) = sum_{k=0}^{oo}(-1)^k{ x^{ 2k+1}}/{(2k+1)!}` .

`1/{1-x} = sum_{k=0}^{oo}x^k` if and only if `| x | < 1`.

**Explanation:** We'll consider only the first of these results.

By the Taylor Theorem applied to the exponential function (Chapter
IX.A), we know
that for any `x`,

`e^x =` exp`(x) = P_n(x) + R_n(x)`
where

`R_n = R_n(x) = e^c {x^{n+1}}/{(n+1)!}` and `c ` is between `0`
and `x`.
To justify the equation as stated we need only show that for any fixed value of `x` as `n->oo` the values of `R_n->0`. Noting that the exponential function is an increasing function , we can say that when `x > 0` , `e^c < e^x` , while when `x < 0` , `e^c < 1 < e^ {|x|}` .

Thus for any `c`, `e^c< e^{|x|}` .

Suppose `|x| < N` and let `B = e^{|x|}{x^ N}/{N!}`
and `r = {|x|}/N`, so `0 < r < 1`.

It is not to hard to show that for
any `n > N` , `|R_n | < B * r^k` where `k = n -
N`.

Now as `n->oo` ,
`k->oo` ,
and since `0 < r < 1`, `r^k->
0`. Thus `R_n-> 0`
also and our argument is done. EOP.

We conclude this section with a result that summarizes the
connection
between the Taylor theory and power series.

**Theorem X.B.2 ( Convergence of Maclaurin - Taylor Series)
**Suppose `f` is a `C^{oo}` function
in an interval containing `x = a`.

Let `P_n(x)= P_n(x,f(x),a) = sum_{k=0}^{k=n} f^ k(a) {(x-a)^k} /{ k!}` and `R_n(x) = f (X) - P_n(x)`.

Then `f (x) = sum_{k=0}^{oo}f^ k(a) {(x-a)^k} /{ k!}` if and only if `R_n(x)->0` as `n->oo`.