Infinite Series: Definition, Examples, Geometric Series, Harmonic Series, Telescoping Sum + MORE

in mathematics •  4 months ago  (edited)

In this video I go over a pretty extensive tutorial on infinite series, its definition, and many examples to elaborate in great detail. An infinite series is the summation of a sequence of terms as the terms approach an infinite number of terms and follow from my earlier video on infinite sequences; thus much of the previous video applies to this video. Several important series are covered such as the Geometric series and the Harmonic Series covered. Also, several interesting derivation techniques are utilized such as partial fraction decomposition and the amazing telescoping sum concept. I go over many examples and 2 important exercises from my calculus book (Early Transcendentals by James Stewart) to make the concept of infinite series, and their divergence or convergence, much more clear.

The topics and sections covered in this video are listed below with their timestamps:

  1. @ 1:34 - Infinite Series and Notation
    • @ 8:29 - Definition 1
  2. @ 13:11 - Example 1: Geometric Series
    • @ 27:43 - Example 2
    • @ 32:49 - Example 3
    • @ 39:49 - Example 4
    • @ 48:28 - Example 5
  3. @ 51:10 - Example 6: Telescoping Sum
  4. @ 1:01:31 - Example 7: Harmonic Series
  5. @ 1:10:30 - Theorem 1
    • @ 1:16:02 - Test for Divergence
    • @ 1:17:30 - Example 8
  6. @ 1:20:35 - Theorem 2
    • @ 1:28:15 - Example 9
  7. End of Chapter Exercises
    • @ 1:35:10 - Exercise 1
    • @ 1:45:43 - Exercise 2

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Infinite Series

Infinite Series.jpeg

Topics to Cover

  1. Infinite Series and Notation
    • Definition 1
  2. Example 1: Geometric Series
    • Example 2
    • Example 3
    • Example 4
    • Example 5
  3. Example 6: Telescoping Sum
  4. Example 7: Harmonic Series
  5. Theorem 1
    • Test for Divergence
    • Example 8
  6. Theorem 2
    • Example 9
  7. End of Chapter Exercises
    • Exercise 1
    • Exercise 2

Infinite Series and Notation

If we try to add the terms of an infinite sequence {an}∞n=1 we get an expression of the form

a1 + a2 + a3 + … + an + …

which is called an infinite series (or just a series) and is denoted, for short, by the symbol

But does it make sense to talk about the sum of infinitely many terms?

It would be impossible to find a finite sum for the series:

1 + 2 + 3 + 4 + 5 + … + n + …

because if we start adding the terms we get the cumulative sums 1, 3, 6, 10, 15, 21, … and, after the n-th term, we get n(n + 1)/2, which becomes very large as n increases.

MES Note: Recall the proof for this formula for the sum of consecutive integers is shown in my earlier video: https://youtu.be/tpkzn2e5mtI

However, if we start to add the terms of the series:

1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + … + 1/2n + …

The following table shows that as we add more and more terms, these partial sums become closer and closer to 1.

MES Note 1: This is similar to my earlier video on Zeno's Paradox in which walking to any given wall is "impossible" since you need to first walk half the distance at each interval thus never reaching it; hence the idea of "limits" arises: http://youtu.be/Uq-8CGkyyqE

MES Note 2: The following Excel spreadsheet file link will be used through this video so make sure to download it and play around with it!

https://1drv.ms/x/s!As32ynv0LoaIh8gwtUIvFFz4HKE0DA

Retrieved: 18 June 2019
Archive: Not Available

In fact, by adding sufficiently many terms of the series we can make the partial sums as close as we like to 1.

So it seems reasonable to say that the sum of this infinite series is 1 and to write:

We use a similar idea to determine whether or not a general series has a sum.

Consider the following partial sums:

s1 = a1
s2 = a1 + a2
s3 = a1 + a2 + a3
s4 = a1 + a2 + a3 + a4

and, in general,

These partial sums form a new sequence {sn}, which may or may not have a limit.

If lim n→∞ sn = s exists (as a finite number), then, as in the preceding example, we call it the sum of the infinite series ∑an.


Definition 1

Given the series ∑n=1 an = a1 + a2 + a3 + …, let sn denote its n-th partial sum:

If the sequence {sn} is convergent and lim n→∞ sn = s exists as a real number, then the series ∑an is called convergent and we write:

The number s is called the sum of the series.

Otherwise, the series is call divergent.


Thus the sum of a series is the limit of the sequence of partial sums.

So when we write ∑n=1 an = s, we mean that by adding sufficiently many terms of the series we can get as close as we like to the number s.

Notice that:

Compare with the improper integral:

To find this integral, we integrate from 1 to t and then let t → ∞.

For a series, we sum from 1 to n and then let n → ∞.


Example 1: Geometric Series

An important example of an infinite series is the geometric series:

Each term is obtained from the preceding one by multiplying it by the common ratio r.

Note that we have already considered the special case where a = 1/2 and r = 1/2 early in this video; i.e. 1/2 + 1/4 + 1/8 + … + 1/2n

If r = 1, then sn = a + a + … + a = na → +/- ∞.

Since limn→∞ sn doesn't exist, the geometric series diverges in this case.

If r ≠ 1, we have:

Note that by multiplying both sides by r we get:

Subtracting these equations, we get:

If -1 < r < 1, recall from Example 10 of my video on Infinite Sequences that rn → 0 as n → ∞.

https://steemit.com/mathematics/@mes/infinite-sequences-limits-squeeze-theorem-fibonacci-sequence-and-golden-ratio-more

Retrieved: 18 June 2019
Archive: http://archive.fo/p1No3

Thus, we have:

Thus when |r| < 1 the geometric series is convergent and its sum is a/(1-r).

If r ≤ -1 or r > 1, the sequence {rn} is divergent (also from the results of the previous video's Example 10), and thus by the above equation, the limit limn→∞ sn does not exist.

Therefore the geometric series diverges in those cases.

We summarize the results of Example 1 as follows.


Summary of Example 1: Geometric Series

The geometric series:

is convergent if |r| < 1 and its sum is:

If |r| ≥ 1, the geometric series is divergent.


Note that in words, the sum of a convergent geometric series is:

First term / (1 - common ratio)

A geometric demonstration of the geometric series is shown below.

If the triangles are constructed as shown and s is the sum of the series, then, by similar triangles (http://youtu.be/gXcQEKVcesc), we get:


Example 2

Find the sum of the geometric series

5 - 10/3 + 20/9 - 40/27 + …

Solution:

The first term is a = 5 and the common ratio is r = - 2/3.

Since |r| = 2/3 < 1, the series is convergent via Example 1 and its sum is:

What do we really mean when we say that the sum of the series in Example 2 is 3?

Of course, we can't literally add an infinite number of terms, one by one.

But, according to Definition 1, the total sum is the limit of the sequence of partial sums.

So, by taking the sum of sufficiently many terms, we can get as close as we like to the number 3.

The table shows the first ten partial sums sn and the graph shows how the sequence of partial sums approaches 3.


Example 3

Is the following series convergent or divergent?

Solution:

Let's rewrite the n-th term of the series in the form arn-1:

We recognize this series as a geometric series with a = 4 and r = 4/3.

Since r > 1, the series diverges as per Example 1.

Note that another way to identify a and r is to write out the first few terms:


Example 4

Write the number 2.3171717… as the ratio of integers.

Solution:

After the first term we have a geometric series with a = 17/103 and r = 1/102.

Therefore:


Example 5

Find the sum of the series:

Solution:

Notice that this series starts with n = 0 and so the first term is x0 = 1.

Important note that with series, we adopt the convention that x0 = 1 even when x = 0.

MES Note: Check out one of my first videos exploring the powers of 0 such as x0 and 00: http://youtu.be/VzFUDiLzRiE

Thus:

This is a geometric series with a = 1 and r = x.

Since |r| = |x| < 1, it converges and from Example 1 gives:


Example 6: Telescoping Sum

Show that the following series is convergent and find its sum.

Solution:

This is not a geometric series, so we go back to the definition of a convergent series and compute the partial sums.

We can simplify this expression if we use partial fraction decomposition (see my earlier videos: https://youtu.be/bqf42x6nZoo.

Thus we have:

Notice that the terms cancel in pairs.

This is an example of a telescoping sum: Because of all the cancellations, the sum collapses (like a pirate's collapsing telescope) into just two terms.

http://www.arsmachina.com/rosstele.htm

Retrieved: 18 June 2019
Archive: http://archive.fo/wrln3

Thus we have:

Therefore the given series is convergent and:

The following figure below illustrates Example 6 by showing the graphs of the sequence of terms an = 1/[n(n+1)] and the sequence {sn} of partial sums.

Notice that an → 0 as sn → 1.

At the end of this video I go over two exercises on two geometric interpretations of Example 6.


Example 7

Show that the harmonic series is divergent.

Solution:

For this particular series it's convenient to consider the partial sums s2 , s4 , s8 , s16 , s32 , … and show that they become large.

This shows that:

Thus {sn} is divergent and therefore the harmonic series diverges.

Note that this method of showing that the harmonic series diverges, according to my calculus book, is due to the French scholar Nicole Oresme (1323-1382).


Theorem 1

If the following series is convergent:

Then:

Proof:

Since ∑an is convergent, the sequence of partial sums {sn} is also convergent.

Let lim n→∞ sn = s.

Since n - 1 → ∞ as n → ∞, we also have limn→∞ sn-1 = s.

Therefore:

Note 1:

  • With any series ∑ an we associate two sequences: the sequence {sn} of its partial sums and the sequence {an} of its terms.

  • If ∑ an is convergent, then the limit of the sequence {sn} is s (the sum of the series) and, as Theorem 1 asserts, the limit of the sequence {an} is 0.

Note 2:

  • The converse of Theorem 1 is not true in general.

  • If lim n→∞ an = 0, we cannot conclude that ∑an is convergent.

  • Observe that for the harmonic series ∑1/n we have an = 1/n → 0 as n → ∞, but we showed in Example 7 that the harmonic sequence ∑1/n is divergent.


The Test for Divergence

If:

does not exist or if:

then the series:

is divergent.


The Test for Divergence follows from Theorem 1 because, if the series is not divergent, then it is convergent, and so limn→∞ an = 0.


Example 8

Show that the following series diverges:

Solution:

So the series diverges by the Test for Divergence; i.e. the sum of the terms keeps getting larger since the terms are not going to 0.

Note 3:

  • If we find the lim n→∞ an ≠ 0, we know that ∑an is divergent.

  • If we find that lim n→∞ an = 0, we know nothing about the convergence or divergence of ∑ an.

  • Remember the warning in Note 2: If limn→∞ an = 0, the series ∑an might converge or it might diverge.


Theorem 2

If ∑an and ∑bn are convergent series, then so are the series ∑ can (where c is a constant), ∑(an + bn ), and ∑(an - bn ), and


These properties of convergent series follow from the corresponding Limit Laws for Sequences from my earlier video on Infinite Sequences.

For instance, here is how part (ii) of Theorem 2 is proved:

Let:

The n-th partial sum for the series ∑(an + bn ) is:

And, taking the limit and breaking the sum into two parts, we have:

Therefore ∑(an + bn ) is convergent and its sum is:


Example 9

Find the sum of the series:

Solution:

The series ∑ 1/2n is a geometric series with a = 1/2 and r = 1/2, so:

In Example 6 we found that:

So, by Theorem 2, the given series is convergent and:

Note 4:

  • A finite number of terms doesn't affect the convergence or divergence of a series.

For instance, suppose that we were able to show that the following series is convergent:

Since:

it follows that the entire series ∑n=1 n/(n3 + 1) is convergent.

Similarly, if it is known that the series ∑n=N+1 an converges, then the full series:

is also convergent.


End of Chapter Exercises

Exercise 1

Graph the curves y = xn, 0 ≤ x ≤ 1, for n = 0, 1, 2, 3, 4, … on a common screen.

By finding the areas between successive curves, give a geometric demonstration of the fact, shown in Example 6, that:

Solution:

Graphing the curves using the amazing Desmos graphing calculator reveals that the total area between the curves from x = 0 to x = 1 approaches the unit square of area 1.

https://www.desmos.com/calculator/ml00mwrvpi

Retrieved: 20 June 2019
Archive: Not Available

The area between successive curves y = xn-1 and y = xn for 0 ≤ x ≤ 1 is:

We can see from the above diagram that as n → ∞, the sum of the areas between the successive curves approaches the area of the unit square, that is 1; thus:


Exercise 2

The below figure shows two circles C and D of radius 1 that touch at P.

T is a common tangent line; C1 is the circle that touches C, D, and T; C2 is the circle that touches C, D, and C1; C3 is the circle that touches C, D, and C2.

This procedure can be continued indefinitely and produces an infinite sequence of circles {Cn}.

Find an expression for the diameter of Cn and thus provide another geometric demonstration of Example 6.

Note that the sum of the diameters of the inner circles are approaching the radius of the larger circles which is equal to 1.

Solution:

Let dn be the diameter of Cn.

We draw lines form the centers of Ci to the center of D (or C), and using the Pythagorean Theorem, we can write:

Recall Difference of Squares (http://youtu.be/qFdohskRQB0:

Thus we have:

Similarly for the second circle:

Likewise for the third circle, we will get:

Thus, in general we have:

If we actually calculate d2 and d3 from the formulas above, we find that they are:

Note that:

Thus we suspect that in general:

To prove this, we use mathematical induction (http://youtu.be/WdIr_onvUtE:

Assume that for all k ≤ n (and recall from the earlier partial fractions decomposition):

Then (and utilizing the telescoping sum from Example 6 as well):

Substituting this into our formula for dk+1, we get:

Thus, the induction is complete since dk+1 is true whenever dk is true.

Now, we observe that the partial sums ∑ni=1 di of the diameters of the circles approach 1 (i.e. the radius of the larger circles) as n → ∞, that is:

Which is what we wanted to prove.

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