Lecture 9: Series (11.2, 11.3)

Series (11.2, 11.3)

Course Syllabus

• Week 1: Revision on integration

• Week 2: Substitution method + tricky problems, By Parts (7.1)

• Week 3: Reduction, Trigonometric Integrals (7.2), Trigonometric Subs. (7.3)

• Week 4: Partial fraction (7.4), Famous Subs, Tricky-problems (7.5)

• Week 5: Arclength (8.1), Area (6.1), Work (6.4)

• Week 6: Surface Area (8.2), Improper integral (7.8)

• Week 7: Volume (6.2, 6.3)

• Week 8: Midterm (25%)

• Week 9: Discuss midterm, Sequence (11.1)

• Week 10: Series (11.2, 11.3)

• Week 11: Series (11.4-11.7)

• Week 12: Power Series (11.8-11.11)

• Week 13: Vectors (12.1-12.4)

• Week 14: Equations of lines and planes (12.5)

Infinite Series

• An infinite series is the sum of an infinite sequence: $S = a<em>1 + a</em>2 + a<em>3 + \cdots + a</em>n + \cdots = \sum a_n$

Example 1

• Suppose we know that the sum of the first n terms of the series $\sum a<em>n$ is $S</em>n = a<em>1 + a</em>2 + \cdots + a_n = \frac{2n}{3n + 5}$

• Then the sum of the series is the limit of the sequence $\lbrace s<em>n \rbrace$: $\sum</em>{n=1}^{\infty} a<em>n = \lim</em>{n \to \infty} S<em>n = \lim</em>{n \to \infty} \frac{2n}{3n + 5} = \lim_{n \to \infty} \frac{2}{3 + \frac{5}{n}} = \frac{2}{3}$

Methods for Determining Convergence/Divergence

• A. Partial sums

• B. Geometric Series

• C. nth term test for divergence

• D. Combined Series

A) Partial Sums

• If $S = \lim<em>{n \to \infty} s</em>n = L$ (single finite value), then the series $S$ is convergent and converges to $L$. Otherwise, it is divergent.

Example 2

• Show that the series $\sum_{n=1}^{\infty} \frac{1}{n(n + 1)}$ is convergent, and find its sum.

Solution:

• Compute the partial sum: $S<em>n = \sum</em>{i=1}^n \frac{1}{i(i + 1)} = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \cdots + \frac{1}{n(n + 1)}$

• Use the partial fraction decomposition: $\frac{1}{i(i + 1)} = \frac{1}{i} - \frac{1}{i + 1}$

• $S<em>n = \sum</em>{i=1}^n \frac{1}{i(i + 1)} = \sum_{i=1}^n (\frac{1}{i} - \frac{1}{i + 1}) = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \cdots + (\frac{1}{n} - \frac{1}{n + 1})$

• $S_n = 1 - \frac{1}{n + 1}$

• $\lim<em>{n \to \infty} S</em>n = \lim_{n \to \infty} (1 - \frac{1}{n + 1}) = 1 - 0 = 1$

B) The Geometric Series

• (fixed ratio between terms)

• $S = a + ar + ar^2 + \cdots + ar^{n-1} + \cdots = \sum_{n=1}^{\infty} ar^{n-1}$

• $s_n = \frac{a(1 - r^n)}{1 - r}$

• $S = \sum<em>{n=1}^{\infty} ar^{n-1} = \lim</em>{n \to \infty} s_n = \frac{a}{1 - r}$ for |r| < 1

Example 3

• Find the sum of the geometric series $5 - 10/3 <a target="_blank" rel="noopener noreferrer nofollow" class="link" href="tel:+ 20/9 - 40/27" data-prevent-progress="true">+ 20/9 - 40/27</a> + \cdots$

Solution

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

• Since |r| = |-2/3| < 1, the series is convergent and its sum is
  $S = \frac{5}{1 - (-2/3)} = \frac{5}{5/3} = 3$

Example 4

• Is the series $\sum_{n=1}^{\infty} 2^{2 \cdot 3^{1-n}}$ convergent or divergent?

Solution

• Rewrite the nth term of the series in the form $ar^{n-1}$:
  $\sum<em>{n=1}^{\infty} 2^{2 \cdot 3^{1-n}} = \sum</em>{n=1}^{\infty} \frac{4^n}{3^{n-1}} = \sum_{n=1}^{\infty} 4 (\frac{4}{3})^{n-1}$

• We recognize this series as a geometric series with $a = 4$ and $r = 4/3$. Since r > 1, the series diverges.

Examples of Convergent Series

• $S = 5 - \frac{10}{3} + \frac{20}{9} - \frac{40}{27} + \cdots$

  • Geometric series: $a = 5$, $r = -\frac{2}{3}$, $S = \frac{5}{1 - (-\frac{2}{3})} = 3$

• $a = \frac{9}{100}$, $r = \frac{1}{100}$

  • Convergent, r < 1

Example 7

• Find the sum of the series $\sum_{n=0}^{\infty} x^n$, where |x| < 1.

• $\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + x^4 + \cdots = \frac{1}{1 - x}$

• If |x| < 1, it converges.

Theorem 6: Properties of Convergent Series

• If the series $\sum a<em>n$ is convergent, then $\lim</em>{n \to \infty} a_n = 0$.

• Proof: Let $S<em>n = a</em>1 + a<em>2 + \cdots + a</em>n$. Then $a<em>n = S</em>n - S<em>{n-1}$. Since $\sum a</em>n$ is convergent, the sequence $\lbrace s<em>n \rbrace$ is convergent. Let $\lim</em>{n \to \infty} S<em>n = s$. Since $n - 1 \to \infty$ as $n \to \infty$, we also have $\lim</em>{n \to \infty} S<em>{n-1} = s$. Therefore, $\lim</em>{n \to \infty} a<em>n = \lim</em>{n \to \infty} (S<em>n - S</em>{n-1}) = \lim<em>{n \to \infty} S</em>n - \lim<em>{n \to \infty} S</em>{n-1} = s - s = 0$

• WARNING: The converse of Theorem 6 is not true in general. If $\lim<em>{n \to \infty} a</em>n = 0$, we cannot conclude that $\sum a<em>n$ is convergent. Observe that for the harmonic series $\sum \frac{1}{n}$, we have $a</em>n = \frac{1}{n} \to 0$ as $n \to \infty$, but $\sum \frac{1}{n}$ is divergent.

Theorem 8: Properties of Convergent Series

• If $\sum a<em>n$ and $\sum b</em>n$ are convergent series, then so are the series $\sum c a<em>n$ (where $c$ is a constant), $\sum (a</em>n + b<em>n)$, and $\sum (a</em>n - b_n)$, and

  • (i) $\sum c a<em>n = c \sum a</em>n$

  • (ii) $\sum (a<em>n + b</em>n) = \sum a<em>n + \sum b</em>n$

  • (iii) $\sum (a<em>n - b</em>n) = \sum a<em>n - \sum b</em>n$

Test for Divergence

• If $\lim<em>{n \to \infty} a</em>n$ does not exist or if $\lim<em>{n \to \infty} a</em>n \neq 0$, then the series $\sum a_n$ is divergent.

Examples

• $\sum_{n=1}^{\infty} 2n$

  • Divergent ($\lim_{n \to \infty} 2n = \infty$)

• $\sum_{n=1}^{\infty} \frac{ln n}{n + 1}$

  • Divergent

Example 10

• Find the sum of the series $\sum_{n=1}^{\infty} (\frac{3}{n(n + 1)} + \frac{1}{2^n})$

Solution

• The series $\sum \frac{1}{2^n}$ is a geometric series with $a = \frac{1}{2}$ and $r = \frac{1}{2}$, so $\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1/2}{1 - 1/2} = 1$

• $\sum<em>{n=1}^{\infty} \frac{3}{n(n + 1)} = 3 \sum</em>{n=1}^{\infty} \frac{1}{n(n + 1)} = 3 \cdot 1 = 3$

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

Combined Series

• Conv. + Conv. = Conv.

• Div. + Div. = Div.

• Div. + Conv. = Div.

• Conv. - Div. = Div.

Example 5: Application

• A drug is administered to a patient at the same time every day. Suppose the concentration of the drug is $C_n$ (measured in mg/mL) after the injection on the nth day. Before the injection the next day, only 30% of the drug remains in the bloodstream, and the daily dose raises the concentration by 0.2 mg/mL.

  • (a) Find the concentration just after the third injection.

  • (b) What is the concentration just after the nth dose?

  • (c) What is the limiting concentration?

Solution

• (a) Just before the daily dose of medication is administered, the concentration is reduced to 30% of the preceding day's concentration, that is, $0.3C<em>n$. With the new dose, the concentration is increased by 0.2 mg/mL, and so $C</em>{n+1} = 0.2 + 0.3C_n$

  • Starting with $C_0 = 0$ and putting $n = 0, 1, 2$ into this equation, we get

    • $C<em>1 = 0.2 + 0.3C</em>0 = 0.2$

    • $C<em>2 = 0.2 + 0.3C</em>1 = 0.2 + 0.2(0.3) = 0.26$

    • $C<em>3 = 0.2 + 0.3C</em>2 = 0.2 + 0.2(0.3) + 0.2(0.3)^2 = 0.278$

  • The concentration after three days is 0.278 mg/mL.

• (b) After the nth dose, the concentration is

  • $C_n = 0.2 + 0.2(0.3) + 0.2(0.3)^2 + \cdots + 0.2(0.3)^{n-1}$

  • This is a finite geometric series with $a = 0.2$ and $r = 0.3$, so

    • $C_n = \frac{0.2[1 - (0.3)^n]}{1 - 0.3} = \frac{2}{7}[1 - (0.3)^n] \text{ mg/mL}$

• (c) Because |0.3| < 1, we know that $\lim_{n \to \infty} (0.3)^n = 0$. So the limiting concentration is

  • $\lim<em>{n \to \infty} C</em>n = \lim_{n \to \infty} \frac{2}{7}[1 - (0.3)^n] = \frac{2}{7}(1 - 0) = \frac{2}{7} \text{ mg/mL}$

P-Series

1. The p-series $\sum \frac{1}{n^p}$ is convergent if p > 1 and divergent if $p \leq 1$.

The Integral Test

• Suppose $f$ is a continuous, positive, decreasing function on $[1, \infty)$ and let $a<em>n = f(n)$. Then the series $\sum a</em>n$ is convergent if and only if the improper integral $\int_1^{\infty} f(x) dx$ is convergent. In other words:

  • (i) If $\int<em>1^{\infty} f(x) dx$ is convergent, then $\sum a</em>n$ is convergent.

  • (ii) If $\int<em>1^{\infty} f(x) dx$ is divergent, then $\sum a</em>n$ is divergent.

When do we use this test?

To use this test, the terms of the series must be:

1. positive, a_n > 0 \forall n.

2. decreasing (nonincreasing), $a<em>{n+1} \leq a</em>n \forall n$.

Example 1

Test the series $\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ for convergence or divergence.

Solution

The function $f(x) = \frac{1}{x^2 + 1}$ is continuous, positive, and decreasing on $[1, \infty)$, so we can use the Integral Test:
$\int<em>1^{\infty} \frac{1}{x^2 + 1} dx = \lim</em>{t \to \infty} \int<em>1^t \frac{1}{x^2 + 1} dx = \lim</em>{t \to \infty} \arctan(x) \Big|<em>1^t = \lim</em>{t \to \infty} (\arctan(t) - \arctan(1)) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$
Thus, $\int<em>1^{\infty} \frac{1}{x^2 + 1} dx$ is a convergent integral and so, by the Integral Test, the series $\sum</em>{n=1}^{\infty} \frac{1}{n^2 + 1}$ is convergent.

NOTE

We should not infer from the Integral Test that the sum of the series is equal to the value of the integral. In fact,
\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}
whereas ∫1∞(1/(𝑥2+1))𝑑𝑥 =𝜋/4
Therefore, in general, ∑𝑎𝑛 ≠∫𝑓(𝑥)𝑑𝑥

Example 4

Determine whether the series ∑(ln 𝑛)/𝑛 (from n=1 to ∞) converges or diverges.
Solution
The function 𝑓(𝑥)=(ln 𝑥)/𝑥 is positive and continuous for 𝑥>1 because the logarithm function is continuous. But it is not obvious whether or not 𝑓 is decreasing, so we compute its derivative:
𝑓′(𝑥) = ((1/𝑥)⋅𝑥−ln 𝑥)/𝑥2 = (1−ln 𝑥)/𝑥2
Thus 𝑓′(𝑥)<0 when ln 𝑥>1, that is, when 𝑥>𝑒. It follows that 𝑓 is decreasing when 𝑥>𝑒 and so we can apply the Integral Test:
∫1∞(ln 𝑥)/𝑥 𝑑𝑥 = lim 𝑡→∞ ∫1𝑡 (ln 𝑥)/𝑥 𝑑𝑥 = lim 𝑡→∞ [(ln 𝑥)2/2]1𝑡 = lim 𝑡→∞ (ln 𝑡)2/2 = ∞
Since this improper integral is divergent, the series ∑(ln 𝑛)/𝑛 is also divergent by the Integral Test.

Remainder Estimate for the Integral Test

2 Remainder Estimate for the Integral Test Suppose 𝑓(𝑘)=𝑎𝑘, where 𝑓 is a continuous, positive, decreasing function for 𝑥≥𝑛 and ∑𝑎𝑛 is convergent. If 𝑅𝑛=𝑠−𝑆𝑛, then
∫𝑛+1∞𝑓(𝑥)𝑑𝑥≤𝑅𝑛≤∫𝑛∞𝑓(𝑥)𝑑𝑥

Examples

EXAMPLE 5
(a) Approximate the sum of the series ∑1/𝑛3 by using the sum of the first 10 terms.
Estimate the error involved in this approximation.
(b) How many terms are required to ensure that the sum is accurate to within 0.0005?
SOLUTION In both parts (a) and (b) we need to know ∫𝑛 ∞ 𝑓(𝑥) 𝑑𝑥. With 𝑓(𝑥)=1/𝑥3,
which satisfies the conditions of the Integral Test, we have
∫𝑛 ∞ 1/𝑥3 𝑑𝑥 = lim 𝑡→∞ ∫𝑛𝑡 1/𝑥3 𝑑𝑥 = lim 𝑡→∞ −1/(2𝑡2) − −1/(2𝑛2) = lim 𝑡→∞ 1/(2𝑛2) − 1/(2𝑡2) = 1/(2𝑛2)
(a) Approximating the sum of the series by the 10th partial sum, we have
∑𝑛=1 ∞ 1/𝑛3 ≈ 𝑆10 = 1/13 + +1/23…1/103 ≈1.1975
According to the remainder estimate in (2), we have
𝑅10≤∫10 ∞ 1/𝑥3 𝑑𝑥 =1/2(10)2 =1/200 = 0.005
So the size of the error is at most 0.005.
(b) Accuracy to within 0.0005 means that we have to find a value of 𝑛 such that
𝑅𝑛≤0.0005. Since
𝑅𝑛≤∫𝑛+1 ∞ 1/𝑥2 𝑑𝑥 =1/2𝑛2
we want
1/2𝑛2<0.0005 Solving this inequality, we get n2 > 1/0.001 = 1000
or
n>√1000 ≈31.6
We need 32 terms to ensure accuracy to within 0.0005.

Estimating the Sum of a Series

So, f ( x ) d x = s = Sn+ + S