[Understanding Analysis by Stephen Abbott] - [Chapter 2 Sequences and Series]

Given an infinite series $\sum_{k=1}^\infty a_k$, it is important to keep a clear distinction between

the sequence of terms: $(a_1, a_2, a_3, ...)$ and
the sequence of partial sums: $(s_1, s_2, s_3,...)$, where $s_n = a_1 + a_2 + \cdot \cdot \cdot + a_n$

The convergence of the series $\sum_{k=1}^\infty a_k$ is defined in terms of the sequence $(s_n)$. Specifically, the statement $$ \sum_{k=1}^\infty a_k = A \text{ means that } \lim s_n = A $$ It is for this reason that we can immediately translate many of our results from the study of sequences into statements about the behavior of infinite series.

Theorem 2.7.1 Algebraic Limit Theorem for Series if $\sum_{k=1}^\infty a_k = A$ and $\sum_{k=1}^\infty b_k = B$, then

$\sum_{k=1}^\infty ca_k = cA$ fo all $c \in R$ and

$\sum_{k=1}^\infty (a_k + b_k) = A + B$

Proof.

In order to show that $\sum_{k=1}^\infty ca_k = cA$, we must argue that the sequence of partial sums $$ t_m = ca_1 + ca_2 + ca_3 + \cdot \cdot \cdot +ca_m $$ converges to $cA$. But we are given that $\sum_{k=1}^\infty a_k$ converges to $A$, meaning that the partial sums converges to $A$. Because $t_m = cs_m$, applying the Algebraic Limit Theorem for sequences yields $(t_m) \to cA$, as desired.

Theorem 2.7.2 Cauchy Criterion for Series. The series $\sum_{k=1}^\infty a_k$ converges if and only if, given $\epsilon > 0$, there exists an $N in N$ such that whenever $n \geqslant m \geqslant N$ it follows that $$ |a_{m+1} + a_{m+2} + \cdot \cdot \cdot + a_n| < \epsilon $$

Proof. Observe that $$ |s_n - s_m| = |a_{m+1} + a_{m+2} + \cdot \cdot \cdot + a_n| $$ and apply the Cauchy Criterion for sequence

Theorem 2.7.3 If the series $\sum_{k=1}^\infty a_k$ converges, then $(a_k) \to 0$

Proof. Consider the special case $n = m + 1$ in the Cauchy Criterion for Series.

Theorem 2.7.4 Comparison Test. Assume $a_k$ and $b_k$ are sequences satisfying $0 \leqslant a_k \leqslant b_k$ for all $k \in N$

if $\sum_{k=1}^\infty b_k$ converges, then $\sum_{k=1}^\infty a_k$ converges.
if $\sum_{k=1}^\infty a_k$ diverges, then $\sum_{k=1}^\infty b_k$ diverges.

Proof. Both statements follow immediately from the Cauchy Criterion for Series and the observation that $$ |a_{m+1} + a_{m+2} + \cdot \cdot \cdot + a_n| \leqslant |b_{m+1} + b_{m+2} + \cdot \cdot \cdot + b_n| $$

Theorem 2.7.6 Absolute Convergence Test. If the series $\sum_{n=1}^\infty |a_n|$ converges, then $\sum_{n=1}^\infty a_n$ converges as well.

Proof. This proof makes use of both the necessity (the "if" direction) and the sufficiency (the "only if" direction) of the Cauchy Criterion for Series. Because $\sum_{n=1}^\infty |a_n|$ converges, we know that, given an $\epsilon > 0$, there exists an $N \in N$ such that $$ |a_{m+1}| + |a_{m+2}| + \cdot \cdot \cdot + |a_n| < \epsilon $$ for all $n > m \geqslant N$. By the triangle inequality $$ |a_{m+1} + a_{m+2} + \cdot \cdot \cdot + a_n| \leqslant |a_{m+1}| + |a_{m+2}| + \cdot \cdot \cdot + |a_n| $$ so the sufficiency of the Cauchy Criterion guarantees that $\sum_{n=1}^\infty a_n$ also converges.

Theorem 2.7.7 Alternating Series Test. Let $(a_n)$ be a sequence satisfying

a_1 \geqslant a_2 \geqslant a_3 \geqslant \cdot \cdot \cdot \geqslant a_n \geqslant a_{n+1} \geqslant \cdot \cdot \cdot and
$(a_n) \to 0$

Definition 2.7.8 if $\sum_{n=1}^\infty |a_n|$ converges, then we say that the original series $\sum_{n=1}^\infty a_n$ converges absolutely. If, on the other hand, the series $\sum_{n=1}^\infty a_n$ converges but the series of the absolute value $\sum_{n=1}^\infty |a_n|$ does not converge, then we say that the original series $\sum_{n=1}^\infty a_n$ converges conditionally.

Definition 2.7.9 Let $\sum_{k=1}^\infty a_k$ be a series. A series $\sum_{k=1}^\infty b_k$ is called a rearrangement of $\sum_{k=1}^\infty a_k$ if there exists a one-to-one, onto function $f : N \to N$ such that $b_{f(k)} = a_k$ for all $k \in N$

Definition 2.7.10 If a series converge absolutely, then any rearrangement of this series converges to the same limit

Proof. Assume $\sum_{k=1}^\infty a_k$ converges absolutely to $A$, and let $\sum_{k=1}^\infty b_k$ be a rearrangement of $\sum_{k=1}^\infty a_k$. Let's use $$ s_n = \sum_{k=1}^n a_k = a_1 + a_2 + \cdot \cdot \cdot + a_n $$ for the partial sums of the original series and use $$ t_m = \sum_{k=1}^n b_k = b_1 + b_2 + \cdot \cdot \cdot + b_m $$ for the partial sums of the rearranged series. Thus we want to show that $(t_m) \to A$.

Let $\epsilon > 0$. By hypothesis, $(s_n) \to A$, so choose $N_1$ such that $$ |s_n -A| < \frac{\epsilon}{2} $$ for all $n \geqslant N_1$. Because the convergence is absolute, we can choose $N_2$ so that $$ \sum_{k=m+1}^n |a_k| < \frac{\epsilon}{2} $$ for all $n > m \geqslant N_2$. Now, take $N = max{N_1, N_2}$. We know that the finite set of terms ${a_1, a_2, a_3,...,a_N}$ must all appear in the rearranged series, and we want to move far enough out in the series $\sum_{n=1}^\infty b_n$ so that we have included all of these terms. Thus choose

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[Understanding Analysis by Stephen Abbott] - [Chapter 2 Sequences and Series] - [2.7 Properties of Infinite Series]