X Sequences and Series: Fundamental Properties (in progress)
© 2000 M. Flashman
March, 2006: This page now requires Internet Explorer 6+MathPlayer or Mozilla/Firefox/Netscape 7+.

X.B Series: Some Key Examples.
X.B.1Geometric Series.

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

`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^3 ... +ar^n+ar^{n+1}`
and then subtract the result from `S_n` to obtain
`S_n = a + ar+ ar^2 + ... +ar^n `
           `- rS_n =`    ` ar+ ar^2 +... +ar^n+ar^{n+1}`  
`(1 - r) S_n = a - ar^{n+1}`                       
and therefore when `r` is not equal to `1`,
`S_n = {a - ar^{ n+1}}/{1-r} = a/{1-r} - a {r^{n+1}}/{1-r}`

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`.



  1. To see a dynamic java sketch that visualizes this result for `0< r < 1`, click here.
  2. 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 )`.
  3. 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}`.
  4. 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.
  5. 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

`a/{1-r} = 3/{1 - .4} = 3/.6 = 5` .

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 .

B.3 The Alternating Harmonic Series

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 subsequence 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` ,

`S_m - S_n = S_{2k} - S_{2k-1} = {(-1)^{n+2}}/(n+1) ->0`
and thus  when `m` is even `S_m ->bar S`.  So whether `n` is an odd or an even number, when `n` is very large, `S_n ~~ bar S`., or as `n ->oo, S_n-> bar S`.

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

`P_n (x, ln(x), 1) = 1 - (x-1) + 1/2 (x-1)^2 - 1/3 (x-1)^3 + ... + (-1)^{n+1}1/n (x-1)^n`.

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

Harmonic Series:

`sum_{k=1}^{oo}1/k` diverges.
`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

`P(x, f ) =sum_{k=0}^{oo} f^k(0) {x^k}/{ k!}` .

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.)

`e^x = sum _{k=0}^{oo} {x^k}/{k!}` .
`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`.