Sequences and Series Notes

Sequences and Series

Introduction to Infinite Sums

  • Consider infinite sums like \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots + \frac{1}{2^i} + \cdots .
  • The question is whether we can assign a numerical value to such an infinite sum.
  • This relates to the idea of a quantity getting "closer and closer" to a fixed quantity, which is assessed by evaluating if the sum approaches a fixed value as more terms are added.
  • Example: \frac{1}{2} = \frac{1}{2} , \frac{1}{2} + \frac{1}{4} = \frac{3}{4} , \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8} , \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} = \frac{15}{16} , etc.
  • It is evident that the sum approaches 1.
  • In fact, it can be shown that \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots + \frac{1}{2^i} = \frac{2^i - 1}{2^i} = 1 - \frac{1}{2^i} .
  • Then, \lim_{i \to \infty} 1 - \frac{1}{2^i} = 1 - 0 = 1 .

Infinite Sums and Series

  • The idea of the infinite sum is used long accepted in representing numbers:
    • For example: 0.3333\overline{3} = \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \frac{3}{10000} + \cdots = \frac{1}{3} .
    • And 3.14159 \ldots = 3 + \frac{1}{10} + \frac{4}{100} + \frac{1}{1000} + \frac{5}{10000} + \frac{9}{100000} + \cdots = \pi .
  • Our primary task is to investigate infinite sums, which are called series by investigating limits of sequences of numbers.
  • \sum_{i=1}^{\infty} \frac{1}{2^i} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots + \frac{1}{2^i} + \cdots is a series.
  • \frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16}, \ldots, \frac{2^i - 1}{2^i}, \ldots is a sequence.
  • The value of a series is defined as the limit of a particular sequence, that is, \sum{i=1}^{\infty} \frac{1}{2^i} = \lim{i \to \infty} \frac{2^i - 1}{2^i} .

Sequences as Functions

  • The idea of a sequence of numbers a1, a2, a_3, \ldots can also be thought of as a function.
  • The sequences are functions whose domains are the natural numbers \mathbb{N} = {1, 2, 3, \ldots} or the non-negative integers \mathbb{Z}_{\geq 0} = {0, 1, 2, 3, \ldots} .
  • The range of the function is still allowed to be the real numbers; in symbols, a sequence is a function f: \mathbb{N} \to \mathbb{R} .
  • Sequences are written in different equivalent ways:
    • a1, a2, a_3, \ldots
    • {an}{n=1}^{\infty}
    • {f(n)}_{n=1}^{\infty}
  • As with functions on the real numbers, most often sequences can be expressed by a formula. Examples include a_i = f(i) = 1 - \frac{1}{2^i} .
  • Other examples:
    • f(i) = \frac{i}{i + 1}
    • f(n) = \frac{1}{2^n}
    • f(n) = \sin(n\pi/6)
    • f(i) = \frac{(i - 1)(i + 2)}{2^i}

Limits of Sequences

  • Given a sequence, we are interested in the limit \lim{i \to \infty} f(i) = \lim{i \to \infty} a_i .
  • This limit is analogous to \lim_{x \to \infty} f(x) where x is a real-valued variable.
  • The only change is that we restrict the input values to be integers.

Definition of a Limit

  • Definition 11.1.1: Suppose that {an}{n=1}^{\infty} is a sequence. We say that \lim{n \to \infty} an = L if for every \epsilon > 0 there is an N > 0 so that whenever n > N , |a_n - L| < \epsilon .
  • If \lim{n \to \infty} an = L we say that the sequence converges, otherwise it diverges.

Connection to Limits of Real Functions

  • If f(i) defines a sequence, and f(x) makes sense for real numbers, and \lim{x \to \infty} f(x) = L , then \lim{i \to \infty} f(i) = L as well.
  • However, the converse is not true: the limit of the sequence may exist even if the limit of the corresponding real function does not.
  • Example:
    • Since \lim{x \to \infty} \frac{1}{x} = 0 , it is also the case that \lim{i \to \infty} \frac{1}{i} = 0 , that is, the numbers 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \ldots get closer and closer to 0.
    • However, if f(n) = \sin(n\pi) , the sequence is \sin(0\pi), \sin(1\pi), \sin(2\pi), \sin(3\pi), \ldots = 0, 0, 0, 0, \ldots , so \lim_{n \to \infty} f(n) = 0 .
    • But \lim{x \to \infty} f(x) = \lim{x \to \infty} \sin(x\pi) does not exist, because as x gets larger and larger, the values \sin(x\pi) do not approach a single value; instead, they take on all values between -1 and 1 over and over.

Computing Limits

  • To compute \lim{n \to \infty} f(n) , try to compute \lim{x \to \infty} f(x) . If the latter exists, it is also equal to the first limit.
  • If \lim{x \to \infty} f(x) does not exist, it may still be true that \lim{n \to \infty} f(n) exists, but another method must be used to compute it.

Graph of a Sequence

  • Since a sequence is defined only for integer values, its graph is a sequence of dots.

Limit Theorems for Sequences

  • The properties of limits of real functions translate easily into properties of sequences.
  • Theorem 11.1.2: Suppose that \lim{n \to \infty} an = L and \lim{n \to \infty} bn = M and k is some constant. Then:
    • \lim{n \to \infty} kan = k \lim{n \to \infty} an = kL
    • \lim{n \to \infty} (an + bn) = \lim{n \to \infty} an + \lim{n \to \infty} b_n = L + M
    • \lim{n \to \infty} (an - bn) = \lim{n \to \infty} an - \lim{n \to \infty} b_n = L - M
    • \lim{n \to \infty} (an bn) = \lim{n \to \infty} an \cdot \lim{n \to \infty} b_n = LM
    • \lim{n \to \infty} \frac{an}{bn} = \frac{\lim{n \to \infty} an}{\lim{n \to \infty} b_n} = \frac{L}{M} , if M is not 0
  • Theorem 11.1.3 (Squeeze Theorem): Suppose that an \leq bn \leq cn for all n > N , for some N . If \lim{n \to \infty} an = \lim{n \to \infty} cn = L , then \lim{n \to \infty} b_n = L .
  • Theorem 11.1.4: \lim{n \to \infty} |an| = 0 if and only if \lim{n \to \infty} an = 0 . (The size of an gets close to zero if and only if an gets close to zero.)

Examples

  • Example 11.1.5: Determine whether {\frac{n}{n + 1}}_{n=0}^{\infty} converges or diverges. If it converges, compute the limit.
    • Since this makes sense for real numbers, we consider \lim{x \to \infty} \frac{x}{x + 1} = \lim{x \to \infty} 1 - \frac{1}{x + 1} = 1 - 0 = 1 .
    • Thus, the sequence converges to 1.
  • Example 11.1.6: Determine whether {\frac{\ln n}{n}}_{n=1}^{\infty} converges or diverges. If it converges, compute the limit.
    • We compute \lim{x \to \infty} \frac{\ln x}{x} = \lim{x \to \infty} \frac{1/x}{1} = 0 , using L’Hôpital’s Rule.
    • Thus, the sequence converges to 0.
  • Example 11.1.7: Determine whether {(-1)^n}_{n=0}^{\infty} converges or diverges. If it converges, compute the limit.
    • This sequence is 1, -1, 1, -1, 1, \ldots and clearly diverges.
  • Example 11.1.8: Determine whether {\left(-\frac{1}{2}\right)^n}_{n=0}^{\infty} converges or diverges. If it converges, compute the limit.
    • We consider the sequence {\left|\left(-\frac{1}{2}\right)^n\right|}{n=0}^{\infty} = {\left(\frac{1}{2}\right)^n}{n=0}^{\infty} .
    • Then \lim{x \to \infty} \left(\frac{1}{2}\right)^x = \lim{x \to \infty} \frac{1}{2^x} = 0 , so by theorem 11.1.4 the sequence converges to 0.
  • Example 11.1.9: Determine whether {\frac{\sin n}{\sqrt{n}}}_{n=1}^{\infty} converges or diverges. If it converges, compute the limit.
    • Since |\sin n| \leq 1 , 0 \leq \left|\frac{\sin n}{\sqrt{n}}\right| \leq \frac{1}{\sqrt{n}} , and we can use theorem 11.1.3 with an = 0 and cn = \frac{1}{\sqrt{n}} .
    • Since \lim{n \to \infty} an = \lim{n \to \infty} cn = 0 , \lim_{n \to \infty} \frac{\sin n}{\sqrt{n}} = 0 and the sequence converges to 0.
  • Example 11.1.10: A particularly common and useful sequence is {r^n}_{n=0}^{\infty} , for various values of r .
    • If r = 1 the sequence converges to 1 since every term is 1; if r = 0 the sequence converges to 0.
    • If r = -1 this is the sequence of example 11.1.7 and diverges.
    • If r > 1 or r < -1 the terms r^n get large without limit, so the sequence diverges.
    • If 0 < r < 1 then the sequence converges to 0.
    • If -1 < r < 0 then |r^n| = |r|^n and 0 < |r| < 1 , so the sequence {|r|^n}{n=0}^{\infty} converges to 0, so also {r^n}{n=0}^{\infty} converges to 0.
    • In summary, {r^n} converges precisely when -1 < r \leq 1 , in which case
      \lim_{n \to \infty} r^n = \begin{cases} 0 & \text{if } -1 < r < 1 \ 1 & \text{if } r = 1 \end{cases} .

Monotonic and Bounded Sequences

  • Sometimes we will not be able to determine the limit of a sequence, but we still would like to know whether it converges.
  • In some cases we can determine this even without being able to compute the limit.
  • A sequence is called increasing (or sometimes strictly increasing) if ai < a{i+1} for all i .
  • It is called non-decreasing (or sometimes increasing) if ai \leq a{i+1} for all i .
  • Similarly, a sequence is decreasing if ai > a{i+1} for all i and non-increasing if ai \geq a{i+1} for all i .
  • If a sequence has any of these properties, it is called monotonic.

Examples

  • Example 11.1.11: The sequence {\frac{2i - 1}{2i}}_{i=1}^{\infty} = \frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16}, \ldots is increasing.
  • The sequence {\frac{n + 1}{n}}_{i=1}^{\infty} = \frac{2}{1}, \frac{3}{2}, \frac{4}{3}, \frac{5}{4}, \ldots is decreasing.
  • A sequence is bounded above if there is some number N such that an \leq N for every n , and bounded below if there is some number N such that an \geq N for every n .
  • If a sequence is bounded above and bounded below, it is bounded.
  • If a sequence {an}{n=0}^{\infty} is increasing or non-decreasing, it is bounded below (by a0 ), and if it is decreasing or non-increasing, it is bounded above (by a0 ).
  • Theorem 11.1.12: If a sequence is bounded and monotonic, then it converges.
    • If a sequence is increasing and bounded, so each term is larger than the one before, yet never larger than some fixed value N , the terms must then get closer and closer to some value between a_0 and N .
    • It need not be N , since N may be a “too-generous” upper bound; the limit will be the smallest number that is above all of the terms a_i .

Examples

  • Example 11.1.13: All of the terms (2i - 1)/2i are less than 2, and the sequence is increasing. As we have seen, the limit of the sequence is 1: 1 is the smallest number that is bigger than all the terms in the sequence.
  • Similarly, all of the terms (n + 1)/n are bigger than 1/2, and the limit is 1: 1 is the largest number that is smaller than the terms of the sequence.
  • We don’t actually need to know that a sequence is monotonic to apply this theorem—it is enough to know that the sequence is “eventually” monotonic, that is, that at some point it becomes increasing or decreasing.
  • Example 11.1.14: Show that {n^{1/n}} converges.
    • First show that this sequence is decreasing, that is, that n^{1/n} > (n+1)^{1/(n+1)} .
    • Consider the real function f(x) = x^{1/x} when x \geq 1 . We can compute the derivative, f'(x) = x^{1/x} \frac{1 - \ln x}{x^2} , and note that when x \geq 3 this is negative.
    • Since the function has negative slope, n^{1/n} > (n + 1)^{1/(n+1)} when n \geq 3 .
    • Since all terms of the sequence are positive, the sequence is decreasing and bounded when n \geq 3 , and so the sequence converges.
  • Example 11.1.15: Show that {\frac{n!}{n^n}} converges.
    • Again, show that the sequence is decreasing, and since each term is positive, the sequence converges.
    • If \frac{a{n+1}}{an} < 1 then a{n+1} < an , which is what we want to know. So we look at \frac{a{n+1}}{an} :
      \frac{a{n+1}}{an} = \frac{\frac{(n + 1)!}{(n + 1)^{n+1}}}{\frac{n!}{n^n}} = \frac{(n + 1)!}{n!} \frac{n^n}{(n + 1)^{n+1}} = \frac{n + 1}{n + 1} \left(\frac{n}{n + 1}\right)^n = \left(\frac{n}{n + 1}\right)^n < 1 .

Series

  • A series is the sum of a sequence: if {an}{n=0}^{\infty} is a sequence, then the associated series is \sum{i=0}^{\infty} an = a0 + a1 + a_2 + \cdots .
  • Associated with a series is a second sequence, called the sequence of partial sums {sn}{n=0}^{\infty} . The partial sum sn is defined as sn = \sum{i=0}^{n} ai . So s0 = a0 , s1 = a0 + a1 , s2 = a0 + a1 + a_2 , etc.
  • A series converges if the sequence of partial sums converges, and otherwise the series diverges.

Geometric Series

  • Example 11.2.1: If an = kx^n , \sum{n=0}^{\infty} a_n is called a geometric series.
  • A typical partial sum is s_n = k + kx + kx^2 + kx^3 + \cdots + kx^n = k(1 + x + x^2 + x^3 + \cdots + x^n) .
  • Note that s_n(1 - x) = k(1 + x + x^2 + x^3 + \cdots + x^n)(1 - x) .
  • Distributing: s_n(1 - x) = k(1 + x + x^2 + x^3 + \cdots + x^n)1 - k(1 + x + x^2 + x^3 + \cdots + x^{n-1} + x^n)x .
  • Then: s_n(1 - x) = k(1 + x + x^2 + x^3 + \cdots + x^n - x - x^2 - x^3 - \cdots - x^n - x^{n+1}) .
  • This simplifies to sn(1 - x) = k(1 - x^{n+1}) , so sn = k \frac{1 - x^{n+1}}{1 - x} .
  • Convergence Condition: If |x| < 1 , \lim{n \to \infty} x^n = 0 , so \lim{n \to \infty} sn = \lim{n \to \infty} k \frac{1 - x^{n+1}}{1 - x} = k \frac{1}{1 - x} .
  • Thus, when |x| < 1 the geometric series converges to \frac{k}{1 - x} .
  • When, for example, k = 1 and x = \frac{1}{2} , then sn = \frac{1 - (1/2)^{n+1}}{1 - 1/2} = \frac{2^{n+1} - 1}{2^n} = 2 - \frac{1}{2^n} and \sum{n=0}^{\infty} \frac{1}{2^n} = \frac{1}{1 - 1/2} = 2 .
  • Returning to the original series, \sum{n=1}^{\infty} \frac{1}{2^n} , each partial sum of this series is 1 less than the corresponding partial sum for the geometric series, so of course the limit is also one less than the value of the geometric series, that is, \sum{n=1}^{\infty} \frac{1}{2^n} = 1 .

Theorems about Series

  • The following theorem follows from theorem 11.1.2.
  • Theorem 11.2.2: Suppose that \sum an and \sum bn are convergent series, and c is a constant. Then:
    • \sum can is convergent and \sum can = c \sum a_n
    • \sum (an + bn) is convergent and \sum (an + bn) = \sum an + \sum bn .
  • The converses of the two parts of this theorem are subtly different.
  • Suppose that \sum an diverges; does \sum can also diverge if c is non-zero? Yes: suppose instead that \sum can converges; then by the theorem, \sum (\frac{1}{c})can converges, but this is the same as \sum an , which by assumption diverges. Hence \sum can also diverges.
  • Now suppose that \sum an and \sum bn diverge; does \sum (an + bn) also diverge? Now the answer is no: Let an = 1 and bn = -1 , so certainly \sum an and \sum bn diverge. But \sum (an + bn) = \sum (1 + -1) = \sum 0 = 0 .
  • Of course, sometimes \sum (an + bn) will also diverge, for example, if an = bn = 1 , then \sum (an + bn) = \sum (1 + 1) = \sum 2 diverges.

Convergence Tests

  • In general, the sequence of partial sums sn is harder to understand and analyze than the sequence of terms an , and it is difficult to determine whether series converge and if so to what.
  • Sometimes things are relatively simple, starting with the following.
  • Theorem 11.2.3: If \sum an converges, then \lim{n \to \infty} a_n = 0 .
  • Proof: Since \sum an converges, \lim{n \to \infty} sn = L and \lim{n \to \infty} s{n-1} = L , because this really says the same thing but “renumbers” the terms. By theorem 11.1.2, \lim{n \to \infty} (sn - s{n-1}) = \lim{n \to \infty} sn - \lim{n \to \infty} s{n-1} = L - L = 0 . But sn - s{n-1} = (a0 + a1 + a2 + \cdots + an) - (a0 + a1 + a2 + \cdots + a{n-1}) = an , so as desired \lim{n \to \infty} a_n = 0 .
  • This theorem presents an easy divergence test: if given a series \sum an the limit \lim{n \to \infty} a_n does not exist or has a value other than zero, the series diverges.
  • Note well that the converse is not true: If \lim{n \to \infty} an = 0 then the series does not necessarily converge.
  • Example 11.2.4: Show that \sum_{n=1}^{\infty} \frac{n}{n + 1} diverges.
    • We compute the limit: \lim_{n \to \infty} \frac{n}{n + 1} = 1 \neq 0 .
  • Example 11.2.5: Show that \sum_{n=1}^{\infty} \frac{1}{n} diverges.
    • Here the theorem does not apply: \lim_{n \to \infty} \frac{1}{n} = 0 , so it looks like perhaps the series converges.
    • Consider:
      1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} > 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{2} + \frac{1}{2}
      1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}
      1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{16} > 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{8} + \frac{1}{16} + \cdots + \frac{1}{16} = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}
    • By swallowing up more and more terms we can always manage to add at least another 1/2 to the sum, and by adding enough of these we can make the partial sums as big as we like.
    • In fact, it’s not hard to see from this pattern that 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^n} > 1 + \frac{n}{2} , so to make sure the sum is over 100, for example, we’d add up terms until we get to around 1/2^{198} , that is, about 4 \cdot 10^{59} terms.
    • This series, \sum (1/n) , is called the harmonic series.

The Integral Test

  • It is generally quite difficult, often impossible, to determine the value of a series exactly.
  • In many cases it is possible at least to determine whether or not the series converges, and so we will spend most of our time on this problem.
  • If all of the terms an in a series are non-negative, then clearly the sequence of partial sums sn is non-decreasing. This means that if we can show that the sequence of partial sums is bounded, the series must converge.
  • We know that if the series converges, the terms an approach zero, but this does not mean that an \geq a_{n+1} for every n . Many useful and interesting series do have this property, however, and they are among the easiest to understand.
  • Example 11.3.1: Show that \sum_{n=1}^{\infty} \frac{1}{n^2} converges.
    • The terms 1/n^2 are positive and decreasing, and since \lim_{x \to \infty} \frac{1}{x^2} = 0 , the terms 1/n^2 approach zero.
    • The goal is to seek an upper bound for all the partial sums, that is, we want to find a number N so that s_n \leq N for every n .
    • The method leverages integration:
      sn = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{n^2} < 1 + \int1^n \frac{1}{x^2} dx < 1 + \int_1^{\infty} \frac{1}{x^2} dx = 1 + 1 = 2 , recalling that we computed this improper integral in section 9.7.
    • Since the sequence of partial sums sn is increasing and bounded above by 2, we know that \lim{n \to \infty} s_n = L \leq 2 , and so the series converges to some number at most 2. In fact, it is possible, though difficult, to show that L = \frac{\pi^2}{6} \approx 1.6 .
  • What goes wrong if we try to apply this technique to \sum \frac{1}{n} ? Here’s the calculation:
    • sn = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} < 1 + \int1^n \frac{1}{x} dx < 1 + \int_1^{\infty} \frac{1}{x} dx = 1 + \infty .
    • The problem is that the improper integral doesn’t converge.
    • Note well that this does not prove that \sum \frac{1}{n} diverges, just that this particular calculation fails to prove that it converges.
  • Example 11.3.2: Consider the following:
    • $$ sn = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} > \int1^{n+1} \frac{1}{x} dx = \