Calculus II
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Positive series

01 Theory

Theory 2

Direct Comparison Test (DCT)

Applicability: Both series are positive: and .

Test Statement: Suppose for large enough . (Meaning: for with some given .) Then:

  • Smaller pushes up bigger:
  • Bigger controls smaller:
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02 Illustration

Example - Direct comparison test: rational functions

Direct comparison test: rational functions

(a)

Choose: and

Check:

Observe: is a convergent geometric series

Therefore: converges by the DCT.

(b)

Choose: and .

Check:

Observe: is a convergent -series

Therefore: converges by the DCT.

(c)

Choose: and

Check: (notice that )

Observe: is a convergent -series

Therefore: converges by the DCT.

(d)

Choose: and

Check:

Observe: is a divergent -series

Therefore: diverges by the DCT.

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03 Theory

Theory 3

Some series can be compared using the DCT after applying certain manipulations and tricks.

For example, consider the series . We suspect convergence because for large . But unfortunately, always, so we cannot apply the DCT.

We could make some ad hoc arguments that do use the DCT, eventually:

Trick Method 1:

  • Observe that for we have . (Check it!)
  • But converges, indeed its value is , which is .
  • So the series converges.

Trick Method 2:

  • Observe that we can change the letter to by starting the new at .
  • Then we have:
  • This last series has terms smaller than so the DCT with (a convergent -series) shows that the original series converges too.

These convoluted arguments suggest that a more general version of Comparison is possible.

Indeed, it is sufficient to compare the relative large-n behavior of the two series. We use the termwise ratios to estimate comparative behavior for increasing .

Limit Comparison Test (LCT)

Applicability: Both series are positive: and .

Test Statement: Suppose that . Then: If , i.e. finite non-zero, then:

Extra - LCT edge cases

If or , we can still draw an inference, but only in one direction:

  • If :
  • If :

Extra - Limit Comparison Test explanation

Suppose and . Then for sufficiently large, we know .

Doing some algebra, we get for large.

If converges, then also converges (constant multiple), and then the DCT implies that converges.

Conversely: we also know that , so for all sufficiently large. Thus if converges, also converges, and by the DCT again converges too.

The cases with or are handled similarly.

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04 Illustration

Example - Limit Comparison Test examples

Limit comparison test examples

(a)

Choose: and .

Compare in the limit:

Observe: is a convergent geometric series

Therefore: converges by the LCT.

(b)

Choose: ,

Compare in the limit:

Observe: is a divergent -series

Therefore: diverges by the LCT.

(c)

Choose: and

Compare in the limit:

Observe: is a converging -series

Therefore: converges by the LCT.

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Alternating series

Videos

Review Videos

Videos, Math Dr. Bob:

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05 Theory

Theory 1

Consider these series:

The absolute values of terms are the same between these series, only the signs of terms change.

The first is a positive series because there are no negative terms.

The second series is the negation of a positive series – the study of such series is equivalent to that of positive series, just add a negative sign everywhere. These signs can be factored out of the series. (For example .)

The third series is an alternating series because the signs alternate in a strict pattern, every other sign being negative.

The fourth series is not alternating, nor is it positive, nor negative: it has a mysterious or unknown pattern of signs.

A series with any negative signs present, call it , converges absolutely when the positive series of absolute values of terms, namely , converges.

THEOREM: Absolute implies ordinary

If a series converges absolutely, then it also converges as it stands.

A series might converge due to the presence of negative terms and yet not converge absolutely:

A series is said to be converge conditionally when the series converges as it stands, but the series produced by inserting absolute values, namely , diverges.

The alternating harmonic series above, , is therefore conditionally convergent. Let us see why it converges. We can group the terms to create new sequences of pairs, each pair being a positive term. This can be done in two ways. The first creates an increasing sequence, the second a decreasing sequence:

Suppose gives the sequence of partial sums of the original series. Then gives the first sequence of pairs, namely , , , . And gives the second sequence of pairs, namely , , , .

The second sequence shows that is bounded above by , so is monotone increasing and bounded above, so it converges. Similarly is monotone decreasing and bounded below, so it converges too, and of course they must converge to the same thing.

The fact that the terms were decreasing in magnitude was an essential ingredient of the argument for convergence. This fact ensured that the parenthetical pairs were positive numbers.

Alternating Series Test (AST) - “Leibniz Test”

Applicability: Alternating series only: with

Test Statement: If:

  1. as (i.e. it passes the SDT: if this fails, conclude diverges)
  2. are decreasing, so

Then:

“Next Term Bound” rule for error of the partial sums:

Extra - Alternating Series Test: Theory

Just as for the alternating harmonic series, we can form positive paired-up series because the terms are decreasing:

The first sequence is monotone increasing from , and the second is decreasing from . The first is therefore also bounded above by . So it converges. Similarly, the second converges. Their difference at any point is which is equal to , and this goes to zero. So the two sequences must converge to the same thing.

By considering these paired-up sequences and the effect of adding each new term one after the other, we obtain the following order relations:

Thus, for any even and any odd :

Now set and subtract from both sides:

Now set and subtract from both sides:

This covers both even cases () and odd cases (). In either case, we have:

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06 Illustration

Example - Alternating Series Test: Basic illustration

Alternating series test: basic illustration

(a) converges by the AST.

Notice that diverges as a -series with .

Therefore the first series converges conditionally.

(b) converges by the AST.

Notice the funny notation: .

This series converges absolutely because , which is a -series with .

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Example - Approximating

Approximating pi

The Taylor series for is given by:

Use this series to approximate with an error less than .

Solution

(1) The main idea is to use and thus . Therefore:

and thus:

(2) Write for the error of the approximation, meaning .

By the AST error formula, we have .

We desire such that . Therefore, calculate such that , and then we will know:

(3) The general term is . Plug in in place of to find . Now solve:

We conclude that at least terms are necessary to be confident (by the error formula) that the approximation of is accurate to within .

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Ratio test and Root test

Videos

Review Videos

Videos, Math Dr. Bob

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07 Theory

Theory 1

Ratio Test (RaT)

Applicability: Any series with nonzero terms.

Test Statement:

Suppose that as .

Then:

Extra - Ratio Test - a deeper look

To understand the ratio test better, first consider this series:

  • The term is created by multiplying the prior term by .
  • The term is created by multiplying the prior term by .
  • The term is created by multiplying the prior term by .

When , the multiplication factor giving the next term is necessarily less than . Therefore, when , the terms shrink faster than those of a geometric series having . Therefore this series converges.

Similarly, consider this series:

Write for the ratio from the prior term to the current term . For this series, .

This ratio falls below when , after which the terms necessarily shrink faster than those of a geometric series with . Therefore this series converges.

The main point of the discussion can be stated like this:

Whenever this is the case, then eventually the ratios are bounded below some , and the series terms are smaller than those of a converging geometric series.

Extra - Ratio Test proof

Let us write for the ratio to the next term from term .

Suppose that as , and that . This means: eventually the ratio of terms is close to ; so eventually it is less than .

More specifically, let us define . This is the point halfway between and . Since , we know that eventually .

Any geometric series with ratio converges. Set for big enough that . Then the terms of our series satisfy , and the series starting from is absolutely convergent by comparison to this geometric series.

(Note that the terms do not affect convergence.)

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08 Illustration

Example - Ratio test

Ratio test examples

(a) Observe that has ratio and thus . Therefore the RaT implies that this series converges.

Simplify the ratio:

Notice this technique! We frequently use these rules:

(To simplify ratios with exponents and factorials.)

(b) has ratio .

Simplify this:

So the series converges absolutely by the ratio test.

(c) Observe that has ratio as .

So the ratio test is inconclusive, even though this series fails the SDT and obviously diverges.

(d) Observe that has ratio as .

So the ratio test is inconclusive, even though the series converges as a -series with .

(e) More generally, the ratio test is usually inconclusive for rational functions; it is more effective to use LCT with a -series.

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09 Theory

Theory 2

Root Test (RooT)

Applicability: Any series.

Test Statement:

Suppose that as .

Then:

Extra - Root test: explanation

The fact that and implies that eventually for all high enough . Set (the midpoint between and ).

Now, the equation is equivalent to the equation .

Therefore, eventually the terms are each less than the corresponding terms of this convergent geometric series:

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10 Illustration

Example - Root test examples

Root test examples

(a) Observe that has roots of terms:

Because , the RooT shows that the series converges absolutely.

(b) Observe that has roots of terms:

Because , the RooT shows that the series converges absolutely.

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Example - Ratio test versus root test

Ratio test versus root test

Determine whether the series converges absolutely or conditionally or diverges.

Solution Before proceeding, rewrite somewhat the general term as .

Now we solve the problem first using the ratio test. By plugging in we see that

So for the ratio we have:

Therefore the series converges absolutely by the ratio test.

Now solve the problem again using the root test. We have for :

To compute the limit as we must use logarithmic limits and L’Hopital’s Rule. So, first take the log:

Then for the first term apply L’Hopital’s Rule:

So the first term goes to zero, and the second (constant) term is the value of the limit. So the log limit is , and the limit (before taking logs) must be (inverting the log using ) and this is . Since , the root test also shows that the series converges absolutely.

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