Series of Real Numbers

definition: An infinite series $\sum_i a_i$ is a pair $\{a_i\},\{s_i\}$ where $a$ is a sequence and $s$ is a sequence of partial sums defined inductively by $s_1 = a_1, s_i = s_{i-1} + a_i$ . Here $a_i$ is the $i$ th term of the series and $s_i$ is the $i$ th partial sum.

definition: $\sum a$ converges to $L$ if $s$ converges to $L$ .

thoerem: if $\sum s$ converges to $A$ and $\sum t$ converges to $B$ then $\sum s + t$ converges to $A + B$ .

theorem: if $\sum a$ is convergent then $a$ is convergent to zero.

$\sum a$ convergent. Then $\lim_n s_{n+1} = L$ and $\lim_n s_n = L$ . So $\lim_n a_{n+1} = (s_{n+1} - s_n) = 0$ .

theorem: if $\sum a$ is such that all terms are non-negative, then $\sum a$ is convergent if $s$ is bounded above and divergent if $s$ is not bounded above.

theorem: if $x \in (0,1)$ then $\sum_n x^n$ is convergent to $(1-x)^{-1}$

Since $s_n = \frac{1 - x^{n+1}}{1-x}$ and since $\lim_n x^n = 0$ , the conclusion follows.

theorem: the series $\sum_n n^{-1}$ is divergent

A proof by induction will suffice.

$s_1 = 1$

$s_2 = 1 + 1/2 = 3/2$

$s_4 = 1 + 1/2 + 1/3 + 1/4 \ge 1 + 1/2 + 1/4 + 1/4 = 2$

$s_8 = s_4 + 1/5 + 1/6 + 1/7 + 1/8 \ge 2 + 4 \cdot 1/8 = 5/2$

By induction, $s_{2^n} = (n+2)/2$ and so contains a divergent sequence. and hence diverges.

definition: an alternating series is a series $\sum a$ such that it can be written as $\sum_n a_n (-1)^{n | n+1}$ where $a$ is a series of non-negative terms.

theorem: if $a$ is a non-increasing sequence of positive numbers convergent to zero, then $\sum_n (-1)^{n+1} a_n$ is convergent.

Note that $s_3 = s_1 - a_2 + a_3 \le s_1$ and in general $\{s_{2n+1}\}$ forms a non-increasing sequence. But $s_{2n-1} = \sum_{i=1..n-1} (a_{2i-1} - a_{2i}) + a_{2n-1} > 0$ . And so the sequence is bounded below and hence convergent. Analogously, $s_{2n}$ is non-decreasing and since $s_{2n} = a_1 - (a_2 - a_3) \ldots - a_{2n} \le a_1$ . and so bounded above and hence convergent.

Let $L = \lim s_{2n}$ and $M = \lim s_{2n+1}$ . Then $0 = \lim_n a_{2n} = \lim_n (s_{2n} - s_{2n-1}) = \lim_n s_{2n} - \lim_n s_{2n-1} = L - M$ and so $L = M$ . Hence the sequence converges to $\{s_n\}$ converges to $L$ .

corollary: if an alternating series satisfies the hypothesis of the theorem above and converges to $L$ then $|s_n - L| \le a_{n+1}$ .

since $s_{2n-1} \ge L$ and $s_{2n} \le L$ and so $0 \le s_{2n-1} - L \le s_{2n-1} - s_{2n} = a_{2n}$ and so $|s_{2n-1} - L| \le a_{2n}$ . Similarly, $0 \le L - s_{2n} \le s_{2n+1} - s_{2n} \le a_{2n+1}$ and so $|s_{2n} - L| \le a_{2n+1}$ .

This corollary may be used to obtain an upper bound on the error, and if a partial sum is computed, an estimate for the limit.

definition: if $\sum |a|$ converges, then $\sum a$ converges absolutely. If $\sum a$ converges but $\sum |a|$ diverges then $\sum a$ converges conditionally.

theorem: if $\sum |a|$ converges then $\sum a$ converges.

It is sufficient to demonstrate $\sum a$ is cauchy. Since $|\sum_{i=m..n} a_i|\le \sum_{i=m..n} |a_i|$ and the series $\sum |a|$ is convergent, it is cauchy and so is $\sum a$ .

definition: the positive part $a^+$ of a sequence is $a_i^+ = \max(a_i,0)$ . the negative part $a^-$ of a sequence is $a^- = -\max(-a_i,0)$ .

Note that $a_i = a_i^+ - a_i^-$ . and $|a_i| = a_i^+ + a_i^-$

theorem: If $\sum a$ converges absolutely, then $\sum a^+$ and $\sum a^-$ converge. However, if $\sum a$ converges conditionally then both diverge.

If $\sum |a|$ is convergent then so is $\sum a$ . But since $\sum a^+ = (a + |a|)/2$ is the sum of convergent series it too is convergent, and so $\sum a^- = \sum (a - a^+)$ is convergent.

If $\sum a$ converges but $\sum |a|$ is divergent, then $|a| + a = 2a^+$ . But if $\sum a^+$ is convergent then so is $\sum |a|$ , which is false.

definition: a rearrangement of $\sum a$ is $\sum a \circ \phi$ where $\phi$ is an isomorphism of the natural numbers.

For example, the series $\sum_n (-1)^{n+1}/n$ is conditionally convergent. Since

$S = +\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\ldots$

$\frac{1}{2}S = +\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\ldots$

$\frac{3}{2}S = +\frac{1}{1} + (-\frac{1}{2}+\frac{1}{2}) + \frac{1}{3} + (-\frac{1}{4}-\frac{1}{4}) + \frac{1}{5} + (-\frac{1}{6} + \frac{1}{6}) + \frac{1}{7} + (-\frac{1}{8}-\frac{1}{8}) + \ldots$

$\frac{3}{2}S = +\frac{1}{1} + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \ldots$

the series on the right of $\frac{3}{2}S$ is (modulo 0 terms if included) a re-arrangement of the series on the right of $S$ and they converge to different sums. It can be shown that a rearrangment can be constructed that converges to any value.

theorem: let $\sum a$ be a conditionally convergent series of real numbes. then for any real number $x$ there is a rearrangment of $\sum a$ convergent to $x$ .

Since $\sum a$ is conditionally convergent, its positive and negative parts are divergent. Let $P(u,v) = \sum_{n=u+1..v} a_{2n-1}$ and $N(u,v) = \sum_{n=u+1..v} a_{2n}$ . Assume $x \ge 0$

Let $m_1$ be the first integer such that $P(0,m_1) > x$ . Let $q_1$ be the first integer such that $P(0,m_1) - N(0,q_1) \le x$ . Let $m_2$ be the first integer such that $P(0,m_1) - N(0,q_1) + P(m_1,m_2) > x$ , and $q_2$ be the first integer such that $P(0,m_1) - N(0,q_1) + P(m_1,m_2) - N(q_1,q_2) \le x$ . This process can be continued, resulting in a rearrangement of $\sum a$ .

To establish convergence, consider the sums

$S_k = P(0,m_1) - N(0,q_1) + \ldots + P(m_{k-1},m_k)$

Then $x < S_k \le x + N(m_{k-1},m_k)$ . Since $\sum a$ is convergent, its terms converge to zero, and so $\lim N(m_{k-1},m_k) = a_{2m_k} \rightarrow 0$ as $k \rightarrow \infty$ and so $S_k$ is convergent to $x$ .

Absolutely convergent series have completely different properties.

lemma: If $\sum a$ is a convergent series of non-negative terms and $b$ is a rearrangement of $a$ , then $\sum b$ is convergent to $\sum a$ .

Let $b = a \circ \psi$ where $\psi$ is a rearrangement of the natural numbers, and let $n_\psi = \max(\psi(i) : i=1..n)$ Let $B_n$ be the $n$ th partial sum of $\sum b$ . Then $B_n \le A_{n_\psi}$ . Hence $\sum b \le \sum a$ . But $a$ is a rearrangement of $b$ and so $\sum a \le \sum b$ .

theorem: if $\sum a$ is absolutely convergent, then any rearrangement $b$ of $a$ is absolutely convergent and $\sum a = \sum b$ .

Since $\sum a$ is absolutely convergent, $\sum a^+$ and $\sum a^-$ are convergent. Let $b$ be a rearrangement of $a$ . Then $b^+$ is a rearrangement of $a^+$ and $b^-$ is a rearrangement of $a^-$ . But then $\sum b^+ = \sum a^+$ and $\sum b^- = \sum a^+$ and so $\sum b = \sum a^+ - \sum a^- = \sum a$ .

definition: the convolution of two sequences $a,b$ indexed at zero is the zero-indexed sequence $a * b$ defined by $(a * b)_n = \sum_{k=0..n} a_k b_{n-k}$ .

This arises naturally when the power series of two sequences is multiplied together and the emergent power series is identified:

$\sum_{i=0..} a_i x^i \sum_{j=0..} b_j x^j = \sum_{i,j=0..} a_i b_j x^{i+j} = \sum_{k = 0..} x^k \sum_{i=0..k} a_i b_{k-i}$

When $x = 1$ , this suggests that $\sum a \sum b = \sum (a * b)$ .

theorem: if $\sum a,\sum b$ are absolutely convergent, then $\sum a * b$ is absolutely convergent and $\sum a \sum b = \sum (a * b)$

Since for any $n$ ,

$\sum_{k=0..n} |(a * b)_k| \le \sum_{k=0..n} \sum_{i=0..k} |a_i| |b_{k-i}| \le \sum_{i=0..n} |a_i| \sum_{i=0..n} |b_i|$

and $\sum_{i=0..n} |a_i| \le \sum |a|$ , $\sum_{i=0..n} |b_i| \le \sum |b|$ , it follows that the partial sums of $\sum |a*b|$ are bounded above and therefore $\sum a* b$ is absolutely convergent. Since any rearrangement of the absolutely convergent series $\sum a* b$ is absolutely convergent to the same sum, it remains to establish that this sum is $\sum a \sum b$ . Note that

$\sum_{k=0..} (a*b)_k = a_0 b_0 + (a_0 b_1 + a_1 b_0) + (a_0 b_2 + a_1 b_1 + a_2 b_0) + \ldots$

Let $C_k = (a * b)_k, A_k = a_0 + \ldots + a_k, B_k = a_0 + \ldots + a_k$ . Then

$C_0 = A_0 B_0$

$C_1 = A_1 B_1 - A_0 B_0$

$C_2 = A_2 B_2 - A_1 B_1$

and in general $C_n = A_n B_n - A_{n-1} B_{n-1}$ .

$C_0 + \ldots + C_n = A_n B_n \rightarrow \sum a \sum b$ as $n \rightarrow \infty$ .

definition: $\sum a$ is dominated by $\sum b$ if $|a|_i \le |b|_i$ for all but finitely many $i$ .

theorem: if $\sum a$ is dominated by an absolutely convergent series $\sum b$ , then $\sum a$ is absolutely convergent.

theorem: if $\sum b$ is dominanted by an an absolutely divergent series $\sum a$ , then $\sum b$ is absolutely divergent.

theorem: if $\sum b$ converges | diverges absolutely and $\lim |a| / |b|$ exists then $\sim a$ converges | diverges absolutely.

Suppose $b$ converges absolutely. Since the ratio is convergent it is bounded by some constant $M$ and so

$\sum |a| = \sum |b| |a| / |b| \le M \sum |b|$

theorem: (ratio test) Let $\sum a$ be a series such that $a \neq 0$ and let $\alpha = \liminf_n |a_{n+1}/a_n|$ and $\beta = \limsup_n |a_{n+1}/a_n|$ .

if $\beta < 1$ then $\sum a$ is absolutely convergent.
if $\alpha > 1$ then $\sum a$ is absolutely divergent.

theorem: (root test) given $\sum a$ let $A(a) = \limsup_n (a_n)^{1/n}$ . Then $\sum a$ converges absolutely when $A(a) < 1$ and diverges when $A(a) > 1$

A useful corollary to the root test is its application to power series

corollary: given $\sum_n a_n x^n$ where $a$ is a real sequence.

if $A(a) = 0$ then the power series is convergent
if $A(a) = L > 0$ then the power series is absolutely convergent when $|x| < L$ and divergent when $|x| > 1/L$ .
if $A(a) = \infty$ , then the power series is convergent only when $x = 0$ .

The tests do not provide information about the convergence of $\sum_n n^{-2}$ . An integral test can be used, but as an alternative that does not use calculus, there is the Cauchy condensation test.

theorem: if $a$ is a non-increasing sequence of positive numbers such that $\sum_n 2^n a_{2^n}$ converges, then $\sum a$ is convergent.

Consider a binary tree with standard node ordering, and populate the nodes of the tree with terms. Since the first node of any row dominates all other nodes in the row, and row $i$ has $2^{i-1}$ nodes, and a filled tree of depth $n$ has $2^n - 1$ nodes

$A_n = \sum_{i=1..2^n - 1} \le \sum_{i=1..n} 2^{i-1} a_{2^i}$

and so $\sum a$ is convergent.

theorem: if $a$ is a non-increasing sequence of positive numbers such that $\sum_n 2^n a_{2^n}$ diverges, then $\sum a$ is divergent.

The proof is similar except instead of using the dominating term on every row of the binary tree, we use the dominated (right side) term.

corollary: $\sum_n 1/n^2$ converges.

since $\sum_n 2^n 1/(2^n)^2 = \sum_n 2^{-n} < \infty$ by the Cauchy Condensaton Test, the series converges.

corollary: $\sum_{n \ge 4} 1/(n \log n)$ diverges.

since $\sum_n 2^n 1/(2^n n) = \sum_n 1/n$ diverges, the series diverges.

A convergent series must approach zero at a rate faster than $1/n$ :

theorem: if $a$ is non-increasing sequence of positive numbers, and $\sum a$ converges then $\lim_n n a_n = 0$ .

Let $S_n = \sum_{i=1..n} a_i$ . If $S = \sum a$ , then any subsequence of $\sum a$ converges to $A$ . Since

$S_{2n} - S_n = a_{n+1} + \ldots a_{2n} \ge n a_{2n} \ge 0$ . and taking the limit $\lim_n n a_{2n} = 0$ . Since $a_{2n+1} \le a_{2n}$ , it follows that

$(2n+1) a_{2n+1} \le (2n+1)/(2n) 2n a_{2n}$ and so $\lim_n (2n+1) a_{2n+1} = 0$ .

There is also a notion called summability by parts.

lemma: let $a,b$ be real sequences. Then $\sum_{i=1..n} a_ib_i = a_{n+1} S_n - \sum_{i=1..n} \Delta a_i \cdot S_i$ .

Let $S_0 = 0, S_i = \sum_{j=1..n} b_j, a_0 = 0$ . Then

$\sum_{i=1..n} a_i b_i = \sum_{i=1..n} a_i \Delta S_{i-1} = \sum_{i=1..n} \Delta (a_{i-1} S_{i-1}) - \Delta a_{i-1} \cdot S_{i-1}$

$= a_n S_n - \sum_{i=1..n} \Delta a_{i-1} \cdot S_{i-1}$

$= a_{n+1} S_n - \Delta a_n \cdot S_n - \sum_{i=1..n} \Delta a_{i-1} \cdot S_{i-1}$

$= a_{n+1} S_n - \sum_{i=1..n} \Delta a_i \cdot S_i$

Since $b_i = \Delta S_{i-1}$ , this can be more suggestively written as

$\sum_{i=1..n} a_i \Delta S_{i-1} = a_{n+1} S_n - \sum_{i=1..n} \cdot S_i \Delta a_i$

which is similar to the integration by parts formula of calculus.

Abel’s lemma: If $a$ is a bounded sequence of real numbers whose partial sums satisfy $m \le A_n \le M$ and if $b$ is non-increasing sequence of non-negative numbers then $mb_1 \le \sum_{i=1..n} a_i b_i \le M b_1$

Since $\sum_{i=1..n} a_i b_i = A_n b_{n+1} + \sum_{i=1..n} A_i (b_i - b_{i+1}) \le M b_1$ and analogously for the lower bound.

theorem: (Dirichlet’s test) Let $a$ be a sequence whose partial sums $A$ are bounded by $M$ and let $b$ be a non-negative non-increasing sequnce convergent to zero. Then $\sum ab$ is convergent.

Let $\epsilon > 0$ be given. For $n \ge m, |\sum_{i=m..n} a_i| = | s_n - s_{m-1}| \le |s_n| + |s_{m-1}| \le 2M$

By Abel’s lemma,

$|\sum_{i=m..n} a_i b_i| \le \sum_{i=m..n}| \le 2M b_m$ . Choose $N$ sufficiently large to ensure that the RHS is smaller than $\epsilon$ when $m \ge N$ . Then the partial sums of $ab$ form a Cauchy sequence and so the series $\sum ab$ is convergent.

theorem: (Abel’s test) Let $\sum a$ be convergent and $b$ be monotone and convergent. Then $\sum ab$ is convergent.

If $b$ is non-decreasing then $c = \lim b - b$ is non-negative and convergent to zero. So $\sum ac$ is convergent. Since $\sum a(\lim b)$ is convergent, it follows that $\sum ab$ is convergent.

Series of Real Numbers

Navigation

About