Real Analysis: Series

Convergent Series

Motivation: Extending Summation to the Infinite Case

Recall the sum of a finite number of elements $\{a_1, \ldots, a_m\}$ using summation notation:

$$\sum\limits_{n=1}^{m} a_n = a_1 + a_2 + \ldots + a_m.$$

What if we would like to extend this to the sum of an infinite number of terms? If $\{a_n\}$ is a real sequence, we would like to come up with some definition for the following notation:

$$\sum\limits_{n=1}^{\infty} a_n = a_1 + a_2 + \ldots.$$

Since this question involves the matter of infinity, it is natural that the answer should involve limits.

Convergent Series

The sum of the first $n$ elements of a real sequence$\{a_n\}$ is called the $n$th partial sum. We can concisely refer to the individual partial sums of a sequence by defining a variable such as $s_n = \sum\limits_{k=1}^{n} a_k.$ An infinite series, or simply a series, is the limit of the sequence of its partial sums:

$$\sum\limits_{n=1}^{\infty} a_n = \lim\limits_{n \rightarrow \infty} s_n.$$

The series $\sum\limits_{n=1}^{\infty} a_n$ is said to converge if the sequence $\{s_n\}$ of its partial sums converges. Formally, we say

$$\sum\limits_{n=1}^{\infty} a_n = L \; \text{ if and only if } \; \lim\limits_{n \rightarrow \infty} s_n = L.$$

A series diverges if it does not converge. Like a divergent sequence, a divergent series may tend towards either positive or negative infinity, switch between several values, oscillate around a value, or devolve into some other pattern.

Algebraic Limit Theorem for Series

Assume $\sum\limits_{n=1}^{\infty} a_n$ and $\sum\limits_{n=1}^{\infty} b_n$ both converge. Then the following properties hold:

  • $\sum\limits_{n=1}^{\infty} ca_n = c\sum\limits_{n=1}^{\infty} a_n$

  • $\sum\limits_{n=1}^{\infty} a_n + b_n = \sum\limits_{n=1}^{\infty} a_n + \sum\limits_{n=1}^{\infty}b_n$

Note: The converse of the second part is not true! That is, the convergence of $\sum\limits_{n=1}^{\infty} a_n + b_n$ does not imply the convergence of either $\sum\limits_{n=1}^{\infty} a_n$ or $\sum\limits_{n=1}^{\infty} b_n.$

Noteworthy Series

There are several series whose convergence properties are important both for proving more general convergence tests of other series, which are covered in the next section, and for broader mathematical purposes. Here are a few:

  • The harmonic series $\sum\limits_{n=1}^{\infty} \dfrac{1}{n}$ diverges.

  • The generalized harmonic series $\sum\limits_{n=1}^{\infty} \dfrac{1}{n^p}$ converges when $p > 1$ and diverges when $p \leq 1.$

  • The geometric series $\sum\limits_{n=1}^{\infty} x^n = \dfrac{x}{x-1}$ when $0 \leq x < 1$ and diverges when $x \geq 1.$ Note that sometimes $0^0$ is defined to be $1,$ in which case $\sum\limits_{n=0}^{\infty} x^n = \dfrac{1}{1-x}.$

Problems

  1. Algebraic Limit Theorem for Series 1: Show that if $\sum\limits_{n=1}^{\infty} a_n$ converges, then $\sum\limits_{n=1}^{\infty} ca_n = c\sum\limits_{n=1}^{\infty}a_n.$

    Consider the partial sums $s_k = \sum\limits_{n=1}^{k} a_n = a_1 + \ldots + a_k$ and $t_k = \sum\limits_{n=1}^{k} ca_n = ca_1 + \ldots + ca_k.$ We can see that $t_k = c(a_1 + \ldots + a_k) = cs_k.$ Applying the algebraic limit theorem gives us the desired result:

    $ c\sum\limits_{n=1}^{\infty} a_n = c\lim\limits_{n \rightarrow \infty} s_n \\ c\sum\limits_{n=1}^{\infty} a_n = \lim\limits_{n \rightarrow \infty} cs_n \\ c\sum\limits_{n=1}^{\infty} a_n = \lim\limits_{n \rightarrow \infty} t_n \\ c\sum\limits_{n=1}^{\infty} a_n = \sum\limits_{n=1}^{\infty} ca_n$

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  2. Algebraic Limit Theorem for Series 2: Show that if $\sum\limits_{n=1}^{\infty} a_n$ and $\sum\limits_{n=1}^{\infty} b_n$ both converge, then $\sum\limits_{n=1}^{\infty} a_n + b_n = \sum\limits_{n=1}^{\infty} a_n + \sum\limits_{n=1}^{\infty} b_n.$

    Let $\{s_n\}$ and $\{t_n\}$ be the sequence of partial sums for $\sum\limits_{n=1}^{\infty} a_n$ and $\sum\limits_{n=1}^{\infty} b_n,$ respectively. The result follows from the algebraic limit theorem for sequences:

    $ \sum\limits_{n=1}^{\infty} a_n + b_n = \lim\limits_{n \rightarrow \infty} s_n + t_n \\ \sum\limits_{n=1}^{\infty} a_n + b_n = \lim\limits_{n \rightarrow \infty} s_n + \lim\limits_{n \rightarrow \infty} t_n \\ \sum\limits_{n=1}^{\infty} a_n + b_n = \sum\limits_{n=1}^{\infty} a_n + \sum\limits_{n=1}^{\infty} b_n $

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  3. Counterexample: Give examples of series $\{a_n\}$ and $\{b_n\}$ that satisfy the following requirements, or explain that none exist:

    1. $\sum\limits_{n=1}^{\infty} a_n + b_n$ converges, but neither $\sum\limits_{n=1}^{\infty} a_n$ nor $\sum\limits_{n=1}^{\infty} b_n$ converges.

    2. $\sum\limits_{n=1}^{\infty} a_n + b_n$ and $\sum\limits_{n=1}^{\infty} a_n$ converge, but $\sum\limits_{n=1}^{\infty} b_n$ diverges.

    3. $\sum\limits_{n=1}^{\infty} a_n + b_n$ diverges, $\sum\limits_{n=1}^{\infty} a_n$ converges, and $\sum\limits_{n=1}^{\infty} b_n$ diverges.

    4. $\sum\limits_{n=1}^{\infty} a_n + b_n$ diverges, but both $\sum\limits_{n=1}^{\infty} a_n$ and $\sum\limits_{n=1}^{\infty} b_n$ converge.

    1. Let $a_n = 1$ and $b_n = -1$ for all $n \in \mathbb{N}$. Then $\sum\limits_{n=1}^{\infty} a_n + b_n = \sum\limits_{n=1}^{\infty} 1 - 1 = \sum\limits_{n=1}^{\infty} 0 = 0.$ But $\sum\limits_{n=1}^{\infty} a_n = \infty$ and $\sum\limits_{n=1}^{\infty} b_n = -\infty,$ thus the two individual series diverge.

    2. This is not possible. Assume $\sum\limits_{n=1}^{\infty} a_n$ converges, $\sum\limits_{n=1}^{\infty} b_n$ diverges, and $\sum\limits_{n=1}^{\infty} a_n + b_n$ converges. By the algebraic limit theorem for series, $\sum\limits_{n=1}^{\infty} -a_n = -\sum\limits_{n=1}^{\infty} a_n.$ By the algebraic limit theorem for series again, we see that $\sum\limits_{n=1}^{\infty} -a_n + \sum\limits_{n=1}^{\infty} a_n + b_n$ also converges. It follows that $\sum\limits_{n=1}^{\infty} -a_n + \sum\limits_{n=1}^{\infty} a_n + b_n = \sum\limits_{n=1}^{\infty} -a_n + (a_n + b_n) = \sum\limits_{n=1}^{\infty} b_n.$ But this implies that $\sum\limits_{n=1}^{\infty}b_n$ converges, which is a contradiction. Thus it cannot be that $\sum\limits_{n=1}^{\infty} a_n + b_n$ converges after all.

    3. Let $a_n = 1$ and $b_n = 0$ for all $n \in \mathbb{N}.$ Then $\sum\limits_{n=1}^{\infty} a_n$ diverges, $\sum\limits_{n=1}^{\infty} b_n$ converges to $0,$ and $\sum\limits_{n=1}^{\infty} a_n + b_n$ diverges.

    4. This is not possible. The Algebraic Limit Theorem shows that the convergence of the individual sums implies the convergence of the combined sum.

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  4. Show that if all terms in a sequence $\{a_n\}$ are non-negative, then the sequence of partial sums $\{s_n\}$ where $s_n = \sum\limits_{k=1}^{n} a_k$ is a monotonically increasing sequence.

    Proof by induction. We show that each term in the sequence is less than its successor.

    Base case: By definition we see that $s_1 = a_1$ and $s_2 = a_1 + a_2.$ Since $a_2$ is non-negative, it follows that $s_2 \geq s_1.$

    Inductive step: Assume $s_{k-1} \leq s_k$. By definition, $s_{k+1} = s_k + a_{k+1}.$ Since $a_{k+1}$ is non-negative, it follows that $s_{k+1} \geq s_k.$ The result follows by induction.

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  5. Show that if a series $\sum\limits_{n=1}^{\infty} a_n$ converges, then $\lim\limits_{n \rightarrow \infty} a_n = 0.$

    Contrapositive: If $\lim\limits_{n \rightarrow \infty} a_n \neq 0,$ then $\sum\limits_{n=1}^{\infty} a_n$ diverges.

    Assume $\sum\limits_{n=1}^{\infty} a_n = p$ for some $p \in \mathbb{R},$ and let $s_n = \sum\limits_{k=1}^n a_k.$ By definition of convergent series, $\lim\limits_{k \rightarrow \infty} s_k = p.$ Then again by definition of sequence convergence, for every $\frac{1}{2}\varepsilon > 0,$ there exists some $N \in \mathbb{N}$ such that $|s_k - p| < \frac{1}{2}\varepsilon$ for every $k \geq N.$ It follows that $|s_{k+1} - p| < \frac{1}{2}\varepsilon$ as well. The triangle inequality shows the following:

    $ |a_{k+1}| = |s_{k+1} - s_k| \\ |a_{k+1}| \leq |s_{k+1} - p| + |p - s_k| \\ |a_{k+1}| < \frac{1}{2}\varepsilon + \frac{1}{2}\varepsilon \\ |a_{k+1}| < \varepsilon $

    Therefore $\lim\limits_{n \rightarrow \infty } a_n = 0.$
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  6. Harmonic series: Show that $\sum\limits_{n = 1}^{\infty} \dfrac{1}{n}$ diverges.

    Hint: Consider powers of 2.

    Informal Overview:

    Consider the first term $a_1 = 1.$ Note that $a_1 > \dfrac{1}{2}.$ Next, note the sum of the next two terms $\dfrac{1}{2} + \dfrac{1}{3} > \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}.$ Once again, note the sum of the next four terms $\dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} > \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} = \dfrac{1}{1}.$ In general, the next $2^n$ terms sum to over $\dfrac{1}{2}.$ We use this fact to show that the sum never converges.

    Formal Proof:

    Define the sums $t_k$ for $k \in \mathbb{N}^+$ as $t_k = \sum\limits_{n = 1}^{2^{k-1}} \dfrac{1}{2^{k-1} + n}.$ Expanding a few terms shows that each $t_k$ is the sum of the next power of two number of terms in the sequence:

    $t_1 = 1 \\ t_2 = \dfrac{1}{2} + \dfrac{1}{3} \\ t_3 = \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} \\ \vdots $

    Note that all the terms in the $k$th sum are greater than $\dfrac{1}{2^k}.$ Since there are $2^{k-1}$ terms in each sum, this means that $t_k > \dfrac{2^{k-1}}{2^k} = \dfrac{1}{2}$ for all $k \in \mathbb{N}^+.$

    Denote the partials sums of $a_n$ as $s_n = \sum\limits_{i = 1}^{n} a_i.$ Note that $s_{2^k} = \sum\limits_{i=1}^{k} t_i > \dfrac{k}{2}.$ It follows that $\lim\limits_{k \rightarrow \infty} s_{2^k} = \infty.$ Since a sequence converges if and only if all of its subequences converge, this means that $\lim\limits_{n \rightarrow \infty} s_n$ also diverges. Therefore by definition of series convergence, $\sum\limits_{n=1}^{\infty} \dfrac{1}{n}$ diverges.

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  7. Counterexample: Show that $\lim\limits_{n \rightarrow \infty} a_n = 0$ does not imply that $\sum\limits_{n=1}^{\infty} a_n$ converges.

    The sequence $\left\{\frac{1}{n}\right\}$ converges to $0$, but series $\sum\limits_{n=1}^{\infty} \frac{1}{n}$ diverges.

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  8. Show that $\sum\limits_{n=1}^{\infty} \dfrac{1}{2^n} = 1.$

    To get a hint at how to prove convergence, let's expand some terms. The first few terms of the sequence of summands are $\dfrac{1}{2}, \dfrac{1}{4}, \dfrac{1}{8},$ and $\dfrac{1}{16}.$ The first few corresponding partial sums are $\dfrac{1}{2}, \dfrac{3}{4}, \dfrac{7}{8},$ and $\dfrac{15}{16}.$ 

    Define the partial sums $s_n$ as $\sum\limits_{k=1}^{n} \dfrac{1}{2^k}.$ We will first prove that $s_n = 1 - \dfrac{1}{2^n}.$

    Proof by induction. Base case: $s_1 = \dfrac{1}{2},$ which in turn is equal to $1 - \dfrac{1}{2^1} = 1 - \dfrac{1}{2} = \dfrac{1}{2}.$

    Inductive step: Assume $s_n = 1 - \dfrac{1}{2^n}.$ Algebra shows the following:

    $s_{n+1} = s_n + \dfrac{1}{2^{n + 1}} \\ s_{n+1} = 1 - \dfrac{1}{2^n} + \dfrac{1}{2^{n+1}} \\ s_{n+1} = 1 - \dfrac{2}{2^{n+1}} + \dfrac{1}{2^{n+1}} \\ s_{n+1} = 1 - \dfrac{1}{2^{n+1}}.$

    The result follows by induction.

    Thus we can rewrite the formula for the partial sums as $s_n = 1 - \dfrac{1}{2^n}.$ Taking the limit gives us $\lim\limits_{n \rightarrow \infty} 1 - \dfrac{1}{2^n} = 1 - \lim\limits_{n \rightarrow \infty} \dfrac{1}{2^n} = 1 - 0 = 1.$ By definition of series convergences, we conclude that $\sum\limits_{k=1}^{\infty} \dfrac{1}{2^k} = 1.$

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  9. Geometric series: Show that if $0^0=1$ and $0 \leq x < 1,$ then $\sum\limits_{n=0}^{\infty} x^n = \dfrac{1}{1-x}.$

    Assume $0 \leq x < 1,$ and consider $\sum\limits_{n=0}^{\infty} x^n.$ Let $\{s_n\}$ be the sequence of partial sums such that $s_n = \sum\limits_{k=0}^{n} x^k.$ Algebra shows the following:

    $s_n = 1 + x + \ldots x^n \\ s_n = \dfrac{(1-x)(1 + x + \ldots x^n)}{1 - x} \\ s_n = \dfrac{(1 + x + \ldots x^n) - (x + x^2 + \ldots x^{n+1})}{1 - x} \\ s_n = \dfrac{1 - x^{n+1}}{1 - x}$

    Taking the limit produces the desired result:

    $ \lim\limits_{n \rightarrow \infty} s_n = \lim\limits_{n \rightarrow \infty} \dfrac{1 - x^{n+1}}{1 - x} \\ \lim\limits_{n \rightarrow \infty} s_n = \dfrac{\lim\limits_{n \rightarrow \infty}(1 - x^{n+1})}{\lim\limits_{n \rightarrow \infty}(1 - x)} \\ \lim\limits_{n \rightarrow \infty} s_n = \dfrac{1 - 0}{1 - x} \\ \lim\limits_{n \rightarrow \infty} s_n = \dfrac{1}{1 - x}$

    Therefore $\sum\limits_{n=0}^{\infty} x^n = \dfrac{1}{1-x}$ when $0 \leq x < 0.$

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