I would like to analyse the following sequences (a, b and c) of families for summability or improper summability:
a) $ a_{i}:=i^{-\alpha} $ for $ \alpha>0 $ :
b) $a_{i}:=(-1)^{i} i^{-\alpha}$ for $\alpha>0 $:
c) $ a_{i}:=\alpha^{i} $ for $ \alpha \in \mathbb{R} $ :
For the purpose of summability, the following applies:
a) The series $ \sum \limits_{i=1}^{\infty} i^{-\alpha} $ is summable for $ \alpha>1 $.
b) The series $ \sum \limits_{i=1}^{\infty}(-1)^{i} i^{-\alpha} $ is summable for $ \alpha>1 $.
c) The series $ \sum \limits_{i=0}^{\infty} \alpha^{i} $ is summable for $ |\alpha|<1 $.
a):
The family $\left(a_{i}\right)_{i \in \mathbb{N}}$ belongs to the general harmonic series. This converges in the case $\alpha>1$ and diverges in the case $\alpha \leq 1$. Therefore, $\left(a_{i}\right)_{i \in \mathbb{N}}$ is summable for $\alpha>1$. Furthermore, $\left(a_{i}\right)_{i \in \mathbb{N}}$ for $ \alpha \leq 1$ is improperly summable, since all numbers $a_{i}, i \in \mathbb{N}$ are positive.
b):
From the definition of summability and subtask a) it immediately follows that $\left(a_{i}\right)_{i \in \mathbb{N}}$ is summable for $\alpha>1$. Now let $\alpha \leq 1$. Since $\left(\left|a_{i}\right|\right)_{i \in \mathbb{N}}$ is a monotonically decreasing zero sequence, the corresponding alternating series converges according to Leibniz' convergence criterion. However, as shown in subtask a), there is no absolute convergence. Consequently, the alternating series converges only conditionally and according to Riemann's rearrangement theorem, a rearrangement exists for each given value in $[-\infty, \infty]$, so that the corresponding series assumes this value. Therefore, $\left(a_{i}\right)_{i \in \mathbb{N}}$ for $\alpha \leq 1$ cannot be improperly summable, but the following must apply $$ \sup \left\{\sum \limits_{i \in E} a_{i}: E \subset \mathbb{N} \text { finally }\right\}=\infty \quad \text { and } \quad \inf \left\{\sum \limits_{i \in E} a_{i}: E \subset \mathbb{N} \text { finally }\right\}=-\infty $$
c):
The family $\left(a_{i}\right)_{i \in \mathbb{N}}$ belongs to the geometric series. Therefore, $\left(a_{i}\right)_{i \in \mathbb{N}}$ is summable if and only if $|\alpha|<1$ applies. Since the numbers $a_{i}, i \in \mathbb{N}$ are all positive for $\alpha \geq 1$, $\left(a_{i}\right)_{i \in \mathbb{N}}$ is not actually summable in this case. For $\alpha \leq-1$, $ \left(a_{i}\right)_{i \in \mathbb{N}}$, on the other hand, is obviously not improperly summable.