I did not post anything for more than one month. I was having some problems. Maybe I lost my aim, maybe lost myself and maybe lost in myself. However, I have made up my mind to go on. I will continue to contribute and will ultimately finish Analysis I. Of course, I will then start to cover Analysis II. However, I am here and will be here for myself and most importantly for my small amount of readers. I put my anxiety and stress aside and cleared my mind. La mort du loup and then birth from ashes like a phoenix. Yes, it is what it is like to exist. Being in the world feels strange somehow, and writing digitally or conventionally is the cure to this. Writing anything. Any ink particule. In particular, writing with a fountain pen. It restricted me to write gibberish, to erase my mistakes and allowed me to think deeply, a little advise so to say.

Absolute Convergence

Remember that for the previous theorems we only used positive series. However, if a series does not satisfy $a_n>0$ $\forall n \geq N$, we still may be able to use previous tests by virtue of the following theorem.

$\textbf{\small Theorem 1.21.:}$ If $\sum_{n=1}^{\infty} |a_n|$ converges, then $\sum_{n=1}^{\infty} a_n$ converges. In other words absolute convergest series is convergent.

$\textbf{\small Proof:}$ Let $\sum_{n=1}^{\infty} |a_n|$ converges and define $a_n = a_n - |a_n| + |a_n|$. Observe that,

$$ \begin{align} 0 \leq |a_n| - a_n < |a_n| + |a_n| &= 2|a_n|\\ |a_n| - a_n &< 2|a_n| \notag \end{align} $$
$$ \begin{align} 0 \leq |a_n| - a_n < |a_n| + |a_n| &= 2|a_n|\\ |a_n| - a_n &< 2|a_n| \notag \end{align} $$

Using direct comparison test with $(1)$ we can say that $\sum_{n=1}^{\infty}|a_n|-a_n$ converges, since $|a_n| - a_n < 2|a_n|$ and $\sum_{n=1}^{\infty} |a_n|$ converges. Lastly, $\sum_{n=1}^{\infty} a_n$ converges, since

$$ \begin{aligned} \sum_{n=1}^{\infty} a_n = \underbrace{\sum_{n=1}^{\infty} (a_n - |a_n|)}_{\text{converges}} + \underbrace{ \sum_{n=1}^{\infty} |a_n|}_{\text{converges}} \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} a_n = \underbrace{\sum_{n=1}^{\infty} (a_n - |a_n|)}_{\text{converges}} + \underbrace{ \sum_{n=1}^{\infty} |a_n|}_{\text{converges}} \end{aligned} $$

Note that, converse of the theorem does not hold. $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ is convergent, but $\sum_{n=1}^{\infty} \left|\frac{(-1)^n}{n}\right| = \sum_{n=1}^{\infty} \frac{1}{n}$ is divergent.

Let us introduce new terminology

$\textbf{\small Definition 1.21.:}$

  1. If $\sum_{n=1}^{\infty} |a_n|$ converges, then $\sum_{n=1}^{\infty} a_n$ converges absolutely
  2. If $\sum_{n=1}^{\infty} a_n$ converges, but $\sum_{n=1}^{\infty} |a_n|$ diverges, then we say that the series $\sum_{n=1}^{\infty} a_n$ converges conditionally

As an example to absolute convergence we may take the series $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$ and to an example of converges conditionally series we may take $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$.

$\textbf{\small Definition 1.22.:}$ (Alternating Series) A series $\sum_{n=1}^{\infty} (-1)^{n-1} a_n$ where $a_n>0$ for all $n \in \mathbb{N}$ is called alternating series.

Of course we have a special convergence test for alternating series. It either called Alternating Series Test or Leibniz’s Test.

$ \textbf{\small Theorem 1.21.:}$ If $(a_n)$ is decreasing and $\lim_{n \rightarrow \infty} a_n = 0$ then the alternating series $\sum_{n=1}^{\infty}(-1)^{n-1}a_n$ converges.

$ \textbf{\small Proof:}$ In order to prove that the alternating series $\sum_{n=1}^{\infty}(-1)^{n-1}a_n$ we define its partial sum

$$ \begin{aligned} s_n=a_1-a_2+a_3-\dots+(-1)^{n-1}a_n \end{aligned} $$
$$ \begin{aligned} s_n=a_1-a_2+a_3-\dots+(-1)^{n-1}a_n \end{aligned} $$

And then we will take subsequence of $s_n$. Rigorously,

$$ \begin{aligned} s_{2n}=a_1-a_2+a_3-\dots+a_{2n-1}-a_{2n} \end{aligned} $$
$$ \begin{aligned} s_{2n}=a_1-a_2+a_3-\dots+a_{2n-1}-a_{2n} \end{aligned} $$

Observe that $s_{2n+2} - s_{2n} = a_{2n+1} - a_{2n+2} \geq 0$ since $(a_n)$ is decreasing sequence. Now let us show $s_{2n}$ is bounded. See that,

$$ \begin{aligned} s_{2n}&=a_1 - \underbrace{(a_2-a_3)}_{\geq 0} - \underbrace{(a_4 - a_5)}_{\geq 0} - \dots - \underbrace{(a_{2n-2} - a_{2n-1})}_{\geq 0} - \underbrace{a_{2n}}_{\geq 0}\\ s_{2n} &< a_1 \end{aligned} $$
$$ \begin{aligned} s_{2n}&=a_1 - \underbrace{(a_2-a_3)}_{\geq 0} - \underbrace{(a_4 - a_5)}_{\geq 0} - \dots - \underbrace{(a_{2n-2} - a_{2n-1})}_{\geq 0} - \underbrace{a_{2n}}_{\geq 0}\\ s_{2n} &< a_1 \end{aligned} $$

$s_{2n}$ is bounded. By monotone convergence theorem we can say $s_{2n}$ converges. Say $\lim_{n\rightarrow \infty} s_{2n} = s$. Now, let us show odd-terms are also convergent. See that

$$ \begin{aligned} s_{2n+1} = s_{2n} + a_{2n+1} \end{aligned} $$
$$ \begin{aligned} s_{2n+1} = s_{2n} + a_{2n+1} \end{aligned} $$

Letting $n\rightarrow \infty$ we have

$$ \begin{aligned} &\lim_{n\rightarrow \infty} s_{2n+1} = \lim_{n\rightarrow \infty} s_{2n} + \lim_{n\rightarrow \infty} a_{2n+1} = s + 0 = s \\ &\lim_{n\rightarrow \infty} s_{2n+1} = s \end{aligned} $$
$$ \begin{aligned} &\lim_{n\rightarrow \infty} s_{2n+1} = \lim_{n\rightarrow \infty} s_{2n} + \lim_{n\rightarrow \infty} a_{2n+1} = s + 0 = s \\ &\lim_{n\rightarrow \infty} s_{2n+1} = s \end{aligned} $$

where $\lim a_{2n+1}=0$ by hypothesis. All in all, even-termed subsequence $s_{2n} \rightarrow s$ and odd-termed subsequence $s_{2n+1} \rightarrow s$. Hence $s_n$ is convergent, which implies $\sum_{n=1}^{\infty}(-1)^{n-1}a_n$ is convergent.

$ \textbf{\small Example:}$ Consider the series

$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n^p} \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n^p} \end{aligned} $$

Let us investigate its convergence. Define $a_n = \frac{1}{n^p}$. $a_n$ is decreasing and $\lim_{n \rightarrow \infty} a_n = 0$ for $p > 0$. By Leibniz’s Test series converges for $p>0$. But for $p = 1$ the series converges conditionally. So

$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n^p} = \begin{cases} \text{absolutely converge} & \text{if } p > 1\\ \text{conditionally converge} & \text{if } 0 < p \leq 1\\ \text{diverges} & \text{if } p \leq 0\\ \end{cases} \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n^p} = \begin{cases} \text{absolutely converge} & \text{if } p > 1\\ \text{conditionally converge} & \text{if } 0 < p \leq 1\\ \text{diverges} & \text{if } p \leq 0\\ \end{cases} \end{aligned} $$

$\textbf{\small Theorem 1.22.:}$ (Error Estimation for Alternating Series) Suppose that alternating series $\sum_{n=1}^{\infty}(-1)^na_n$ satisfies the condition of Leibniz’s Test. Let $s = \sum_{n=1}^{\infty}(-1)^{n-1}a_n$ and $s_n = a_1 - a_2 + \dots + (-1)^{n-1}a_n$. Then, $|s-s_n| = |r_n| \leq a_{n+1}$.

$\textbf{\small Proof:}$

$$ \begin{aligned} |s - s_n| &= \left|\sum_{n=1}^{\infty} (-1)^{n-1} a_n - \sum_{k=1}^{n} (-1)^{k-1} a_k\right|\\ &=\left| \sum_{k=n+1}^{\infty} (-1)^{k-1} a_k\right|\\ &=\left|a_{n+1} - a_{n+2} + a_{n+3} - a_{n+4} + \dots \right|\\ &=\left|\underbrace{(a_{n+1} - a_{n+2})}_{\geq 0} + \underbrace{(a_{n+3} - a_{n+4})}_{\geq 0} + \dots \right|\\ &=a_{n+1} - \underbrace{(a_{n+2} - a_{n+3})}_{\geq 0} - \underbrace{(a_{n+4} - a_{n+5})}_{\geq 0} - \dots\\ &< a_{n+1} \end{aligned} $$
$$ \begin{aligned} |s - s_n| &= \left|\sum_{n=1}^{\infty} (-1)^{n-1} a_n - \sum_{k=1}^{n} (-1)^{k-1} a_k\right|\\ &=\left| \sum_{k=n+1}^{\infty} (-1)^{k-1} a_k\right|\\ &=\left|a_{n+1} - a_{n+2} + a_{n+3} - a_{n+4} + \dots \right|\\ &=\left|\underbrace{(a_{n+1} - a_{n+2})}_{\geq 0} + \underbrace{(a_{n+3} - a_{n+4})}_{\geq 0} + \dots \right|\\ &=a_{n+1} - \underbrace{(a_{n+2} - a_{n+3})}_{\geq 0} - \underbrace{(a_{n+4} - a_{n+5})}_{\geq 0} - \dots\\ &< a_{n+1} \end{aligned} $$

$\textbf{\small Example:}$ How many terms of the series $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{n}{2^n}$ are needed to compute the sum of the series with error less than 0,001.

Observe, $(a_n) = \left(\frac{n}{2^n}\right)$ is decreasing and $\lim_{n \rightarrow \infty} a_n = 0$. Then, conditions of Leibniz’s Test satisfied given series is convergent. Let

$$ \begin{aligned} s = \sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{n}{2^n} \end{aligned} $$
$$ \begin{aligned} s = \sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{n}{2^n} \end{aligned} $$

Then the difference

$$ \begin{aligned} |s - s_n| < a_{n+1} =\dfrac{n+1}{2^{n+1}} < 0,001 \end{aligned} $$
$$ \begin{aligned} |s - s_n| < a_{n+1} =\dfrac{n+1}{2^{n+1}} < 0,001 \end{aligned} $$

Hence, $n = 13$.

Rearrangements of Series

$\textbf{\small Definition 1.23.:}$ (Rearrangement) Let $\sum_{n=1}^{\infty} a_n$ be a series. A series $\sum_{n=1}^{\infty}b_k$ is called a rearrangement of $\sum_{n=1}^{\infty} a_n$ if there exists a one-to-one, onto function $f : \mathbb{N} \rightarrow \mathbb{N}$ such that $b_{f(n)} = a_n$ for all $n \in \mathbb{N}$.

$\textbf{\small Theorem 1.23.:}$ (Dirichlet’s Theorem) Any rearrangements of an absolute convergence series converges to the same sum.

$\textbf{\small Proof:}$ (By Stephen Abbot’s Understanding Analysis) Assume $\sum_{n=1}^{\infty}a_n$ converges to $A$, and let $\sum_{n=1}^{\infty} b_n$ be a rearrangement of $\sum_{n=1}^{\infty}a_n$. Let

$$ \begin{aligned} s_n = a_1 + a_2 + \dots + a_n\\ t_m = b_1 + b_2 + \dots + b_m \end{aligned} $$
$$ \begin{aligned} s_n = a_1 + a_2 + \dots + a_n\\ t_m = b_1 + b_2 + \dots + b_m \end{aligned} $$

Thus is enough to show $\lim_{m \rightarrow \infty} t_m = A$. Let $\varepsilon > 0$. Choose $N_1$ such that

$$ \begin{aligned} |s_n - A| < \dfrac{\varepsilon}{2} \end{aligned} $$
$$ \begin{aligned} |s_n - A| < \dfrac{\varepsilon}{2} \end{aligned} $$

for all $n \geq N_1$. Since the convergence is absolute we can choose $N_2$ such that

$$ \begin{aligned} \sum_{k=m+1}^{n}|a_k| < \dfrac{\varepsilon}{2} \end{aligned} $$
$$ \begin{aligned} \sum_{k=m+1}^{n}|a_k| < \dfrac{\varepsilon}{2} \end{aligned} $$

for all $n > m > N_2$. Take $N = \max \{N_1, N_2\}$. We know that the finite set of terms $\{a_1, a_2, \dots, a_N\}$ must all appear in the rearranged series. Also, we must choose $M$ so that, in the series $\sum_{n=1}^{\infty} b_n$ all the terms are included. Choose $M = \max \{f(k) : 1 \leq k \leq N\}$.

Now, see that if $m \geq M$, then $(t_m - s_N)$ consists of a finite set of terms. Our choice of $N_2$ then guarantees $|t_m -s_N|<\frac{\varepsilon}{2}$, and so

$$ \begin{aligned} |t_m - A| &= |t_m - s_N + s_N - A|\\ &\leq |t_m - s_N| + |s_N - A|\\ &< \dfrac{\varepsilon}{2} \end{aligned} $$
$$ \begin{aligned} |t_m - A| &= |t_m - s_N + s_N - A|\\ &\leq |t_m - s_N| + |s_N - A|\\ &< \dfrac{\varepsilon}{2} \end{aligned} $$

whenever $m \geq M$.