Week 1 notes

Week 1 Continuity of the inverse function. Infinite series

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These notes are being developed to reflect the content of the Real Analysis course as taught in the 2024/25 academic year. The first half of the course is lectured by Dr Yuri Bazlov. Questions and comments on these lecture notes should be directed to Yuri.Bazlov@manchester.ac.uk. The second half will be lectured by Dr Mark Coleman.

Pre-requisite: Mathematical Foundations and Analysis

We build upon what was achieved in the Mathematical Foundations and Analysis (MFA) course, taught in Semester 1. Limit of a sequence, continuous function and limit of a function remain key notions in Real Analysis, which develops the “analysis” part of MFA further. Important functions of real variable, used in MFA, will be formally defined, and their properties proved. This includes the power function $x^{𝛼}$ with arbitrary real $𝛼;$ exponential function $a^{x}$ and the logarithm; trigonometric functions.

Hence, the Real Analysis course will constantly require the students to call on their knowledge of definitions and results introduced in MFA.

An informal introduction (not covered in lectures)

The first part of the course will introduce infinite series. Recall Euler’s identity

$e^{i 𝜋} + 1 = 0,$

often brought up as an example of a beautiful mathematical result. But what exactly is there to prove? Let us try to understand this equation and its connection to Real Analysis.

First of all, $e$ is the irrational number discovered by Jacob Bernoulli in 1683 (and initially written as “ $b$ ”) in his study of compound interest. It is defined as

$e = lim_{n \to \infty} (1 + \frac{1}{n})^{n},$ equivalently, $e = \sum_{n = 0}^{\infty} \frac{1}{n!} = 1 + \frac{1}{1} + \frac{1}{1 \cdot 2} + \frac{1}{1 \cdot 2 \cdot 3} + \dots .$

Both formulae involve an infinite process, which in practice can only approximate $e$ to a given precision. Note how the limit of a sequence, defined in MFA, appears next to something new: a “sum of infinitely many numbers”.

What is $e^{i 𝜋},$ though? Raising $e$ to an imaginary power is done via the rule

$e^{z} = 1 + \frac{z}{1!} + \frac{z^{2}}{2!} + \frac{z^{3}}{3!} + \dots$

This is where the meaning of Euler’s identity starts to come across. The “infinite sum” for $e^{i 𝜋}$ breaks down into the real and imaginary parts, leading to two equations,

$1 - \frac{𝜋^{2}}{2!} + \frac{𝜋^{4}}{4!} - \frac{𝜋^{6}}{6!} + \dots = - 1$ and $𝜋 - \frac{𝜋^{3}}{3!} + \frac{𝜋^{5}}{5!} - \frac{𝜋^{7}}{7!} + \dots = 0.$

Adding up ten terms of each “infinite sum”, we obtain approximately $- 1.000000004$ and $- 5.3 \times 10^{- 10}$ (check the calculation here!) which suggests that if we keep adding new terms generated by the same rule, in the limit we will indeed get $- 1$ and $0,$ respectively. This is mysterious: $𝜋$ is transcendental, so no finite sum like this can be a rational number. A proper way to prove that $𝜋$ satisfies these equations is to express $\cos (x)$ and $\sin (x),$ functions given by ratios of sides in a right-angled triangle where $x$ is in radians, as

$\cos (x) = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \dots,$ $\sin (x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \dots$

These infinite expansions, apparently known already to Madhava (ca. 1400), are enormously important in mathematics and have many compelling applications in the sciences. To prove these formulas is the same as to prove the formula $e^{i x} = \cos (x) + i \sin (x)$ (used in MFA without proof). Complete and rigorous theory, leading to these expansions, is part of what is covered in the Real Analysis course. In the first part of MATH11112, we will:

• formally define infinite series and make sense of “sums of infinitely many numbers”;
• learn about ways to tell whether a given series converges, i.e., has a sum;
• understand power series which consist of power of $x$ with some coefficients, and see why they define “smooth” continuous functions of $x$ if they converge.

Expansion of functions as power series is intimately connected with differentiation, a formal treatment of which begins the second part of the course. Higher derivatives of a function are key to approximating the function by polynomials, called Taylor polynomials. It is this theory that allows us to write a “good” function as sum of a Taylor series.

The course concludes with the third part devoted to rigorous treatment of integration. A key result is the Fundamental Theorem of Calculus, which demonstrates that integration is truly a reverse operation to differentiation.

Further study of series: a power series is a sum of infinitely many functions of the form $a x^{n} .$ In 1807, Joseph Fourier publicised a class of scientific problems which require calculating infinite sums of more sophisticated functions, such as $a \sin (n x) .$ The theory of series that we develop in Real Analysis serves as a foundation for the study of Fourier series and other advanced series in mathematics, science and engineering.

End of the informal introduction.

The Inverse Function Theorem

We begin with a result which could have been proved in Mathematical Foundations and Analysis (MFA), given more time. The Inverse Function Theorem will be used to define a function $x \mapsto x^{r}$ where $r$ is rational. First, a definition.

Definition.

Let $A \subseteq R .$ A real-valued function $f$ on $A$ is

.
increasing,	if	$\forall x, x^{'} \in A,$ $x < x^{'}$ $⟹$ $f (x) \leq f (x^{'});$
strictly increasing,	if	$\forall x, x^{'} \in A,$ $x < x^{'}$ $⟹$ $f (x) < f (x^{'});$
decreasing,	if	$\forall x, x^{'} \in A,$ $x < x^{'}$ $⟹$ $f (x) \geq f (x^{'});$
strictly decreasing,	if	$\forall x, x^{'} \in A,$ $x < x^{'}$ $⟹$ $f (x) > f (x^{'}) .$

A function satisfying one of the above conditions is called (strictly) monotone.

The above applies to sequences which are functions on $A = N .$

We can use monotone sequences to calculate limits of functions. The following is an MFA-style result:

Lemma 1.1: limit of $f$ from the left via strictly increasing sequences.

For a function $f,$ defined on $(a, b),$ the following are equivalent:

(i) $lim_{x \to b -} f (x) = ℓ .$
(ii) For all strictly increasing sequences $(x_{n})$ such that $x_{n} \to b$ as $n \to \infty,$ one has $lim_{n \to \infty} f (x_{n}) = ℓ .$

Remark: the Lemma can be expressed in words as follows:

The limit of $f$ at $b$ from the left is the common limit of all sequences $(f (x_{n}))_{n \geq 1},$ where a sequence $(x_{n})_{n \geq 1}$ is strictly increasing and converges to $b .$

Proof of the Lemma (not given in class). (i) $\Rightarrow$ (ii): let $𝜀 > 0$ be arbitrary. First, we use the definition of $lim_{x \to b -} f (x) = ℓ$ to generate $𝛿 > 0$ such that $| f (x) - ℓ | < 𝜀$ for all $x \in (b - 𝛿, b) .$

Now we let $(x_{n})_{n \geq 1}$ be a strictly increasing sequence, and use the above $𝛿 > 0$ in the definition of “ $x_{n} \to b$ as $n \to \infty$ ” to generate $N \in N$ such that $n \geq N$ implies $| x_{n} - b | < 𝛿 .$ That is, $x_{N}, x_{N + 1}, \dots \in (b - 𝛿, b + 𝛿) .$ Since the sequence is strictly increasing with limit $b,$ no term can exceed $b,$ so in fact $x_{N}, x_{N + 1}, \dots \in (b - 𝛿, b) .$

But then, by the choice of $𝛿,$ $| f (x_{n}) - ℓ | < 𝜀$ for all $n \geq N .$ We have shown that $ℓ$ satisfies the definition of limit for the sequence $(f (x_{n}))_{n \geq 1},$ and so (ii) is proved.

(ii) $\Rightarrow$ (i): to prove the contrapositive of this implication, we assume that the statement “ $lim_{x \to b -} f (x) = ℓ$ ” is false. This means that there exists some $𝜀_{0} > 0$ such that for all $𝛿 > 0,$ the interval $(b - 𝛿, b)$ contains a point, say $x (𝛿),$ with $| f (x (𝛿)) - ℓ | \geq 𝜀_{0} .$

Choose $𝛿_{1} = 1$ and construct $x (𝛿_{1}) \in (b - 𝛿_{1}, b) .$ We have $| f (x (𝛿_{1})) - ℓ | \geq 𝜀_{0} .$

Then, for each $n \geq 2,$ choose $𝛿_{n} = min (\frac{1}{n}, b - x (𝛿_{n - 1}))$ and construct $x (𝛿_{n}) \in (b - 𝛿_{n}, b) .$

Since $b - \frac{1}{n} < x (𝛿_{n}) < b,$ by the Sandwich Rule $x (𝛿_{n}) \to b$ as $n \to \infty .$ Also, since $x (𝛿_{n}) > b - (b - x (𝛿_{n - 1})) = x (𝛿_{n - 1}),$ the sequence $x (𝛿_{1}), x (𝛿_{2}), \dots$ is strictly increasing.

We still have, by construction, that $| f (x (𝛿_{n})) - ℓ | \geq 𝜀_{0}$ for all $n .$ Therefore, $ℓ$ fails to satisfy the definition of the limit of the sequence $(f (x (𝛿_{n})))_{n \geq 1},$ i.e. (ii) is false. □

(dots on a graph of a function converging to a limit)

The Lemma is illustrated by Figure 1.1. The next result mirrors the Lemma to deal with a limit from the right:

Corollary: limit of $f$ from the right via strictly decreasing sequences.

For a function $f,$ defined on $(a, b),$ the following are equivalent:

(i) $lim_{x \to a +} f (x) = ℓ .$
(ii) For all strictly decreasing sequences $(x_{n})$ such that $x_{n} \to a$ as $n \to \infty,$ one has $lim_{n \to \infty} f (x_{n}) = ℓ .$ □

In other words,

the limit of $f$ at $a$ from the right is the common limit of all sequences $(f (x_{n}))_{n \geq 1},$ where a sequence $(x_{n})_{n \geq 1}$ is strictly decreasing and converges to $a .$

We are now ready to prove the first theorem of the course.

Theorem 1.2: the Inverse Function Theorem for strictly increasing functions.

A strictly increasing continuous function $f : [a, b] \to [f (a), f (b)]$ has an inverse $g : [f (a), f (b)] \to [a, b]$ which is strictly increasing and continuous.

Proof. $f : [a, b] \to [f (a), f (b)]$ is surjective, because for every $d \in [f (a), f (b)]$ the Intermediate Value Theorem (and its corollary in MFA) gives $c$ in $[a, b]$ such that $f (c) = d .$

A strictly increasing $f$ is injective: indeed, if $x_{1} \neq x_{2},$ then either $x_{1} < x_{2}$ and so $f (x_{1}) < f (x_{2}),$ or $x_{1} > x_{2}$ and so $f (x_{1}) > f (x_{2}) .$ In either case $f (x_{1}) \neq f (x_{2}) .$

We have shown that $f$ is bijective, hence it has an inverse $g = f^{- 1} : [f (a), f (b)] \to [a, b] .$

We prove that $g$ is strictly increasing by contradiction. Assume not, then there exist $y_{1},$ $y_{2}$ such that $y_{1} < y_{2}$ and $g (y_{1}) \geq g (y_{2}) .$ Since $f$ is increasing, $f (g (y_{1})) \geq f (g (y_{2})) .$ Since $f = g^{- 1},$ this reads $y_{1} \geq y_{2},$ but at the same time $y_{1} < y_{2},$ a contradiction.

We prove that $g$ is continuous at an arbitrary point $d$ of its domain by verifying the criterion of continuity, seen in MFA:

$lim_{y \to d} g (y) = g (d) .$

By results from MFA, this is equivalent to

$lim_{y \to d -} g (y) = lim_{y \to d +} g (y) = g (d) .$

We first show that $lim_{y \to d -} g (y) = g (d) .$ We would like to use Lemma 1.1 for this, so we let $(y_{n})$ be a strictly increasing sequence in $[f (a), f (b)]$ which converges to $d .$ Since $g$ is a strictly increasing function, $(g (y_{n}))$ is a strictly increasing sequence; it is also bounded (lies in $[a, b]$ ), hence by a result from MFA, $(g (y_{n}))$ has a limit, say $c,$ in $[a, b] .$

Then by Lemma 1.1, $lim_{n \to \infty} f (g (y_{n})) = lim_{x \to c -} f (x)$ which is $f (c)$ as $f$ is continuous. Since $f = g^{- 1},$ this says that $lim_{n \to \infty} y_{n} = f (c) .$ Thus, $d = f (c),$ hence $g (d) = c .$

We have proved that the common limit of all sequences $(g (y_{n})),$ where $y_{n}$ strictly increases and converges to $d,$ is $g (d) .$ By Lemma 1.1, this means that $lim_{y \to d -} g (y) = g (d) .$

The proof that $lim_{y \to d +} g (y) = g (d)$ is completely similar, based on the Corollary to Lemma 1.1, and we omit it. Continuity of $g$ at $d$ is proved. □

Example: the $p$ ^th root function.

Let $p \in N .$ Show that the $p$ ^th power function $[0, + \infty) \to [0, + \infty),$ $x \mapsto x^{p},$ has a continuous inverse (denoted $y \mapsto \sqrt[p]{y}$ and called the $p$ th root function).

Solution: define $f : [0, + \infty) \to [0, + \infty)$ by $f (x) = x^{p} .$ Then $f$ is strictly increasing on $[0, + \infty) .$ Apply Inverse Function Theorem 1.2 to the restriction $[0, b] \overset{f}{\to} [0, b^{p}]$ to get a continuous inverse $\sqrt[p]{} : [0, b^{p}] \to [0, b] .$ Since $b > 0$ can be made arbitrarily large, this defines the continuous function $\sqrt[p]{}$ on all of $[0, + \infty) .$

We compose continuous, strictly increasing functions to define a rational power function:

Example: raising to rational power $\frac{p}{q}$ where $p \in Z,$ $q \in N$ .

Define $x^{\frac{p}{q}} = {(\sqrt[q]{x})}^{p} .$ This is a continuous function of $x$ where $x \in (0, + \infty) .$

Remark: one can deduce from the definition of a rational power that

$x > 1, r, s \in Q, r < s ⟹ x^{r} < x^{s} .$

This allows us to formally define arbitrary real powers of $x :$

Definition: $x^{𝛼}$ where $x > 0$ and $𝛼 \in R$ .

If $x \geq 1,$ define $x^{𝛼} = sup {x^{r} : r \in Q, r \leq 𝛼} .$

If $0 < x < 1,$ define $x^{𝛼} = (1 / x)^{- 𝛼} .$

A disadvantage of this definition is that proving the expected properties of powers such as $x^{𝛼} x^{𝛽} = x^{𝛼 + 𝛽}$ requires work. We will soon obtain a more useful expression for powers via the exponential function.

Infinite series: definition

Definition: infinite series, convergent series, sum.

For real numbers $a_{n},$ an infinite series is an expression of the form $\sum_{n = 1}^{\infty} a_{n}$ (also written as $a_{1} + a_{2} + \dots + a_{n} + \dots,$ $\sum_{n \geq 1} a_{n}$ or just $\sum a_{n}$ ).

The $n$ th partial sum of this series is the finite sum of terms up to and including $a_{n} :$ $s_{n} = a_{1} + \dots + a_{n} = \sum_{i = 1}^{n} a_{i} .$

If the sequence of partial sums converges: $lim_{n \to \infty} s_{n} = s,$ we say that the series $\sum_{n = 1}^{\infty} a_{n}$ is convergent with sum $s$ , and write $\sum_{n = 1}^{\infty} a_{n} = s .$

Remarks on the definition: (i) $\sum_{n = 1}^{\infty} a_{n} = s$ is a actually a shorthand which means “the series $\sum_{n = 1}^{\infty} a_{n}$ is convergent with sum $s$ ”.

(ii) Any series that is not convergent is said to be a divergent series.

(iii) A series can start from $n = N$ (any integer) instead of $n = 1 :$ $a_{N} + a_{N + 1} + \dots = \sum_{n = N}^{\infty} a_{n} .$ For example, it is common to start from $n = 0.$ The $n$ th partial sum will still be the sum which ends with $a_{n} :$ e.g., for the series $a_{0} + a_{1} + a_{2} + a_{3} + \dots,$

• $a_{0}$ is the $0$ ^th partial sum,
• $a_{0} + a_{1}$ is the $1$ ^st partial sum, $a_{0} + a_{1} + a_{2}$ is the $2$ ^nd partial sum, and so on.

Basic examples of convergent/divergent series are discussed in week 1 supervision classes.

Alert: a strict definition of convergence.

The definition of convergence of the series $\sum_{n = 1}^{\infty} a_{n},$ used in Real Analysis, is very strict: if the number sequence $a_{1},$ $a_{1} + a_{2},$ $a_{1} + a_{2} + a_{3}, \dots$ does not have a limit, then the series has no sum.

Weaker definitions can assign a “sum” to some particular types of series which we consider divergent: Cesàro sum, Abel sum etc. They are used in specialist applications which are beyond this course.

The geometric series

The next example is simple yet important: we will see that more complicated series can be studied by comparing them to a geometric series. We revisit a result seen in MFA.

Proposition 1.3: convergence and sum of geometric series.

Let $a, r \in R .$ The geometric series with initial term $a$ and ratio $r,$

$a + a r + a r^{2} + a r^{3} + \dots = \sum_{n = 0}^{\infty} a r^{n},$

is convergent if $| r | < 1,$ with sum $\frac{a}{1 - r} .$

Proof. The $n$ th partial sum of the series is $s_{n} = a (1 + r + r^{2} + \dots + r^{n}) .$ The calculation

$\begin{aligned} (1 + r + r^{2} + \dots + r^{n}) (1 - r) \\ = 1 - r + r - r^{2} + r^{2} - r^{3} + \dots + r^{n} - r^{n + 1} \\ = 1 - r^{n + 1} \end{aligned}$ where the intermediate terms cancel, gives us the formula

$s_{n} = a \frac{1 - r^{n + 1}}{1 - r} .$

If $| r | < 1,$ we recall from MFA that $r^{n + 1}$ tends to $0$ as $n \to \infty,$ so by Algebra of Limits of convergent sequences,

$s = lim_{n \to \infty} s_{n} = a \frac{1 - 0}{1 - r} = \frac{a}{1 - r} .$

The sum of the series, is, by definition, the limit of partial sums if it exists. Hence the sum of the geometric series is $a / (1 - r)$ as claimed. □

Convergence of series with non-negative terms

Unlike the geometric series, usually there is no nice formula for the $n$ ^th partial sum $s_{n} .$ We still want to decide if a series is convergent, so we prove theorems known as “convergence tests”. Our first few tests work for series where all terms are non-negative.

Theorem 1.4: boundedness test for non-negative series.

Let a series $a_{1} + a_{2} + \dots$ have $a_{n} \geq 0$ for all $n .$ The following are equivalent:

(i) the partial sums $s_{1}, s_{2}, \dots$ are bounded above;
(ii) the series is convergent.

If (i) and (ii) hold, the sum of the non-negative series is the least upper bound, $sup {s_{n} : n \geq 1},$ of its partial sums.

Proof. The partial sums of a non-negative series form an increasing sequence, because $s_{n + 1} = s_{n} + a_{n + 1} \geq s_{n}$ for all $n .$ We know from MFA that an increasing sequence $(s_{n})_{n \geq 1}$ of real numbers has a limit iff it is bounded above, and then the limit is the supremum of the terms of the sequence. □

Corollary: only two convergence types for non-negative series.

A non-negative series $a_{1} + a_{2} + \dots$ is either

• convergent with a non-negative finite sum: $\sum_{n = 1}^{\infty} a_{n} = s,$ $0 \leq s < + \infty,$ or
• divergent if $sup {s_{n} : n \geq 1} = + \infty .$

In the latter case we use the symbolic notation “ $\sum_{n = 1}^{\infty} a_{n} = + \infty$ ” and say that the non-negative series diverges to $+ \infty$ .

Alert.

Notation $\sum_{n = 1}^{\infty} a_{n} = + \infty$ (to mean that the series is divergent) and $\sum_{n = 1}^{\infty} a_{n} < + \infty$ (to mean that the series is convergent) is used only for series with non-negative terms!

The next test is used very often.

Theorem 1.5: the comparison test for non-negative series.

Assume that $0 \leq a_{n} \leq b_{n}$ for all $n .$ If the series $b_{1} + b_{2} + \dots$ is convergent (with sum $T$ ), then the series $a_{1} + a_{2} + \dots$ is convergent (with sum at most $T$ ).

If $a_{1} + a_{2} + \dots$ is divergent, then $b_{1} + b_{2} + \dots$ is also divergent.

Proof. Write $s_{n} = a_{1} + \dots + a_{n}$ and $t_{n} = b_{1} + \dots + b_{n} .$ As $a_{1} \leq b_{1},$ $a_{2} \leq b_{2}$ etc, we have $s_{n} \leq t_{n},$ where $t_{n} \leq T$ by Theorem 1.4. Hence $s_{1}, s_{2}, \dots$ have an upper bound $T,$ so by boundedness test, Theorem 1.4, the series $a_{1} + a_{2} + \dots$ is convergent.

The sum $\sum_{n = 1}^{\infty} a_{n}$ is the least upper bound of $(s_{n})_{n \geq 1},$ and $T$ is an upper bound. Hence $\sum_{n = 1}^{\infty} a_{n} \leq T .$

Now “ $a_{1} + a_{2} + \dots$ is divergent $\Rightarrow$ $b_{1} + b_{2} + \dots$ is divergent” follows by contrapositive. □

To use the Comparison Test, we need to compare with some easy series $\sum b_{n},$ yet the Test does not tell us how to find it. Hence we develop further convergence tests.

Theorem 1.6: the Ratio Test for positive series.

For a positive series $\sum_{n \geq 1} a_{n},$ suppose that the limit $ℓ = lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}}$ exists. Then if $0 \leq ℓ < 1,$ the series is convergent, and if $ℓ > 1,$ the series is divergent.

Proof. The case $0 \leq ℓ < 1.$ Choose a positive $𝜀$ such that $ℓ + 𝜀 < 1.$ For example, $𝜀 = (1 - ℓ) / 2$ works.

Since $ℓ = lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}},$ there exists $N$ such that $\frac{a_{n + 1}}{a_{n}} < ℓ + 𝜀$ for all $n \geq N .$ Write this as $a_{n + 1} < r a_{n},$ where $r = ℓ + 𝜀 .$ Then $a_{N + 2} < r a_{N + 1} < r^{2} a_{N},$ and, repeating this, we obtain $a_{N + k} < r^{k} a_{N} .$ We have

$\begin{aligned} a_{1} + a_{2} + \dots + a_{N + k} & \leq a_{1} + \dots + a_{N - 1} + a_{N} + r a_{N} + \dots + r^{k} a_{N} \\ \leq (a_{1} + \dots + a_{N - 1}) + \frac{a_{N}}{1 - r} . \end{aligned}$ The upper bound that we have obtained is a finite constant which does not depend on $k .$ Thus, partial sums of the series $a_{1} + a_{2} + \dots$ are bounded, so by Theorem 1.4, the series is convergent.

The case $ℓ > 1.$ Put $𝜀 = ℓ - 1.$ There is $N$ such that $ℓ - 𝜀 < \frac{a_{n + 1}}{a_{n}}$ for $n \geq N .$ But $ℓ - 𝜀 = 1,$ so $1 < \frac{a_{n + 1}}{a_{n}},$ equivalently $a_{n} < a_{n + 1},$ for $n \geq N .$ In particular, all $a_{n}$ for $n > N$ are greater than the positive constant $a_{N} .$ Hence $s_{n} \geq (n - N) a_{N}$ which is unbounded. □

Alert: the Ratio test may be inconclusive.

If $ℓ = 1$ or the limit does not exist, this test does not tell us anything: the harmonic series $\sum_{n \geq 1} \frac{1}{n}$ is divergent (see the next Chapter), and the series of inverse squares $\sum_{n \geq 1} \frac{1}{n^{2}}$ is convergent (see the first exercise sheet). Both have $ℓ = 1.$

In the following test (not taught in lectures, not examinable), we use the $n$ th root function $\sqrt[n]{},$ defined earlier.

Theorem 1.7: The $n$ th Root Test for non-negative series.

For a non-negative series $\sum_{n \geq 1} a_{n},$ suppose that the limit $ℓ = lim_{n \to \infty} \sqrt[n]{a_{n}}$ exists. Then if $0 \leq ℓ < 1,$ the series is convergent, and if $ℓ > 1,$ the series is divergent.

Remark: Again, if $ℓ = 1$ or the limit does not exist, this test does not tell us anything.

Proof. (not given in class: very similar to the proof of the Ratio Test; not examinable.) The case $0 \leq ℓ < 1.$ Choose a positive $𝜀$ so that $r = ℓ + 𝜀$ is still less than $1;$ for example, $𝜀 = (1 - ℓ) / 2$ works.

By definition of limit, there is $N$ such that $\sqrt[n]{a_{n}} < ℓ + 𝜀 = r$ for all $n \geq N .$ Then $a_{n} < r^{n}$ for $n \geq N,$ so partial sums of the series are bounded by $a_{1} + \dots + a_{N} + \frac{1}{1 - r},$ implying convergence.

The case $ℓ > 1.$ Put $𝜀 = ℓ - 1.$ Since $ℓ$ is the limit of $\sqrt[n]{a_{n}},$ there is $N$ such that $ℓ - 𝜀 < \sqrt[n]{a_{n}}$ for $n \geq N .$ But $ℓ - 𝜀 = 1,$ so $1 < \sqrt[n]{a_{n}},$ equivalently $1 < a_{n},$ for $n \geq N .$ We therefore have $s_{n} \geq n - N$ which is unbounded. □

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