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, $\sum<em>{i=1}^{\infty} \frac{1}{2^i} = \lim</em>{i \to \infty} \frac{2^i - 1}{2^i}$.

Sequences as Functions

• The idea of a sequence of numbers $a<em>1, a</em>2, a_3, \ldots$ 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:
  • $a<em>1, a</em>2, a_3, \ldots$
  • ${a<em>n}</em>{n=1}^{\infty}$
  • ${f(n)}_{n=1}^{\infty}$
• 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 $\lim<em>{i \to \infty} f(i) = \lim</em>{i \to \infty} a_i$.
• 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 ${a<em>n}</em>{n=1}^{\infty}$ is a sequence. We say that $\lim<em>{n \to \infty} a</em>n = L$ if for every \epsilon > 0 there is an N > 0 so that whenever n > N , |a_n - L| < \epsilon .
• If $\lim<em>{n \to \infty} a</em>n = L$ 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 $\lim<em>{x \to \infty} f(x) = L$, then $\lim</em>{i \to \infty} f(i) = L$ 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 $\lim<em>{x \to \infty} \frac{1}{x} = 0$, it is also the case that $\lim</em>{i \to \infty} \frac{1}{i} = 0$, 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 $\lim<em>{x \to \infty} f(x) = \lim</em>{x \to \infty} \sin(x\pi)$ 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 $\lim<em>{n \to \infty} f(n)$, try to compute $\lim</em>{x \to \infty} f(x)$. If the latter exists, it is also equal to the first limit.
• If $\lim<em>{x \to \infty} f(x)$ does not exist, it may still be true that $\lim</em>{n \to \infty} f(n)$ 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 $\lim<em>{n \to \infty} a</em>n = L$ and $\lim<em>{n \to \infty} b</em>n = M$ and $k$ is some constant. Then:
  • $\lim<em>{n \to \infty} ka</em>n = k \lim<em>{n \to \infty} a</em>n = kL$
  • $\lim<em>{n \to \infty} (a</em>n + b<em>n) = \lim</em>{n \to \infty} a<em>n + \lim</em>{n \to \infty} b_n = L + M$
  • $\lim<em>{n \to \infty} (a</em>n - b<em>n) = \lim</em>{n \to \infty} a<em>n - \lim</em>{n \to \infty} b_n = L - M$
  • $\lim<em>{n \to \infty} (a</em>n b<em>n) = \lim</em>{n \to \infty} a<em>n \cdot \lim</em>{n \to \infty} b_n = LM$
  • $\lim<em>{n \to \infty} \frac{a</em>n}{b<em>n} = \frac{\lim</em>{n \to \infty} a<em>n}{\lim</em>{n \to \infty} b_n} = \frac{L}{M}$, if $M$ is not 0
• Theorem 11.1.3 (Squeeze Theorem): Suppose that $a<em>n \leq b</em>n \leq c<em>n$ for all $n > N$, for some $N$. If $\lim</em>{n \to \infty} a<em>n = \lim</em>{n \to \infty} c<em>n = L$, then $\lim</em>{n \to \infty} b_n = L$.
• Theorem 11.1.4: $\lim<em>{n \to \infty} |a</em>n| = 0$ if and only if $\lim<em>{n \to \infty} a</em>n = 0$. (The size of $a<em>n$ gets close to zero if and only if $a</em>n$ 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 $\lim<em>{x \to \infty} \frac{x}{x + 1} = \lim</em>{x \to \infty} 1 - \frac{1}{x + 1} = 1 - 0 = 1$.
  • 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 $\lim<em>{x \to \infty} \frac{\ln x}{x} = \lim</em>{x \to \infty} \frac{1/x}{1} = 0$, 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 ${\left|\left(-\frac{1}{2}\right)^n\right|}<em>{n=0}^{\infty} = {\left(\frac{1}{2}\right)^n}</em>{n=0}^{\infty}$.
  • Then $\lim<em>{x \to \infty} \left(\frac{1}{2}\right)^x = \lim</em>{x \to \infty} \frac{1}{2^x} = 0$, 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 $a<em>n = 0$ and $c</em>n = \frac{1}{\sqrt{n}}$.
  • Since $\lim<em>{n \to \infty} a</em>n = \lim<em>{n \to \infty} c</em>n = 0$, $\lim_{n \to \infty} \frac{\sin n}{\sqrt{n}} = 0$ 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 ${|r|^n}<em>{n=0}^{\infty}$ converges to 0, so also ${r^n}</em>{n=0}^{\infty}$ converges to 0.
  • In summary, ${r^n}$ converges precisely when -1 < r \leq 1 , in which case
    $\lim_{n \to \infty} r^n = \begin{cases} 0 &amp; \text{if } -1 &lt; r &lt; 1 \ 1 &amp; \text{if } r = 1 \end{cases}$.

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 $a<em>i < a</em>{i+1}$ for all $i$.
• It is called non-decreasing (or sometimes increasing) if $a<em>i \leq a</em>{i+1}$ for all $i$.
• Similarly, a sequence is decreasing if $a<em>i > a</em>{i+1}$ for all $i$ and non-increasing if $a<em>i \geq a</em>{i+1}$ 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 $a<em>n \leq N$ for every $n$, and bounded below if there is some number $N$ such that $a</em>n \geq N$ for every $n$.
• If a sequence is bounded above and bounded below, it is bounded.
• If a sequence ${a<em>n}</em>{n=0}^{\infty}$ is increasing or non-decreasing, it is bounded below (by $a<em>0$), and if it is decreasing or non-increasing, it is bounded above (by $a</em>0$).
• 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 $a<em>{n+1} < a</em>n$, which is what we want to know. So we look at $\frac{a<em>{n+1}}{a</em>n}$:
    \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 ${a<em>n}</em>{n=0}^{\infty}$ is a sequence, then the associated series is $\sum<em>{i=0}^{\infty} a</em>n = a<em>0 + a</em>1 + a_2 + \cdots$.
• Associated with a series is a second sequence, called the sequence of partial sums ${s<em>n}</em>{n=0}^{\infty}$. The partial sum $s<em>n$ is defined as $s</em>n = \sum<em>{i=0}^{n} a</em>i$. So $s<em>0 = a</em>0$, $s<em>1 = a</em>0 + a<em>1$, $s</em>2 = a<em>0 + a</em>1 + a_2$, etc.
• A series converges if the sequence of partial sums converges, and otherwise the series diverges.

Geometric Series

• Example 11.2.1: If $a<em>n = kx^n$, $\sum</em>{n=0}^{\infty} a_n$ 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 $s<em>n(1 - x) = k(1 - x^{n+1})$, so $s</em>n = k \frac{1 - x^{n+1}}{1 - x}$.
• Convergence Condition: If |x| < 1 , $\lim<em>{n \to \infty} x^n = 0$, so $\lim</em>{n \to \infty} s<em>n = \lim</em>{n \to \infty} k \frac{1 - x^{n+1}}{1 - x} = k \frac{1}{1 - x}$.
• Thus, when |x| < 1 the geometric series converges to $\frac{k}{1 - x}$.
• When, for example, $k = 1$ and $x = \frac{1}{2}$, then $s<em>n = \frac{1 - (1/2)^{n+1}}{1 - 1/2} = \frac{2^{n+1} - 1}{2^n} = 2 - \frac{1}{2^n}$ and $\sum</em>{n=0}^{\infty} \frac{1}{2^n} = \frac{1}{1 - 1/2} = 2$.
• Returning to the original series, $\sum<em>{n=1}^{\infty} \frac{1}{2^n}$, 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, $\sum</em>{n=1}^{\infty} \frac{1}{2^n} = 1$.

Theorems about Series

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

Convergence Tests

• In general, the sequence of partial sums $s<em>n$ is harder to understand and analyze than the sequence of terms $a</em>n$, 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 $\sum a<em>n$ converges, then $\lim</em>{n \to \infty} a_n = 0$.
• Proof: Since $\sum a<em>n$ converges, $\lim</em>{n \to \infty} s<em>n = L$ and $\lim</em>{n \to \infty} s<em>{n-1} = L$, because this really says the same thing but “renumbers” the terms. By theorem 11.1.2, $\lim</em>{n \to \infty} (s<em>n - s</em>{n-1}) = \lim<em>{n \to \infty} s</em>n - \lim<em>{n \to \infty} s</em>{n-1} = L - L = 0$. But $s<em>n - s</em>{n-1} = (a<em>0 + a</em>1 + a<em>2 + \cdots + a</em>n) - (a<em>0 + a</em>1 + a<em>2 + \cdots + a</em>{n-1}) = a<em>n$, so as desired $\lim</em>{n \to \infty} a_n = 0$.
• This theorem presents an easy divergence test: if given a series $\sum a<em>n$ the limit $\lim</em>{n \to \infty} a_n$ does not exist or has a value other than zero, the series diverges.
• Note well that the converse is not true: If $\lim<em>{n \to \infty} a</em>n = 0$ 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 $a<em>n$ in a series are non-negative, then clearly the sequence of partial sums $s</em>n$ 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 $a<em>n$ approach zero, but this does not mean that $a</em>n \geq a_{n+1}$ 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 $s<em>n$ is increasing and bounded above by 2, we know that $\lim</em>{n \to \infty} s_n = L \leq 2$, 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 = \