Geometric and Harmonic Series

Example 1.

1. For $q\in \mathbb{K}$, the geometric series $$\sum _{k=0}^{\mathrm{\infty }}{q}^k$$ is convergent if and only if $|q|<1$.

   Proof: We can show that the $n$-th partial sum is given by $$s_n=\sum _{k=0}^n{q}^k=\{\begin{array}{ll}\frac{1-q^{n+1}}{1-q}& :\text{ if }q\ne 1,\\ n+1& :\text{ if }q=1.\end{array}$$ Hence, $(s_n{)}_{n\in \mathbb{N}}$ is convergent if and only if $|q|<1$. In this case we have $$\sum _{k=0}^{\mathrm{\infty }}{q}^k=\underset{n\to \mathrm{\infty }}{lim}{s}_n=\underset{n\to \mathrm{\infty }}{lim}\frac{1-q^{n+1}}{1-q}=\frac{1}{1-q}.$$

2. The harmonic series $$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}$$ is divergent to $+\mathrm{\infty }$.
   Proof: If we construct some unbounded subsequence $(s_{n_l}{)}_{l\in \mathbb{N}}$, the divergence of the harmonic series is proven (since it is monotonically increasing). Indeed, we now show the unboundedness of the subsequence $(s_{2^l}{)}_{l\in \mathbb{N}}$: First, observe that $$s_{2^l}=s_1+(s_2-s_1)+(s_4-s_2)+(s_8-s_4)+\dots +(s_{2^l}-s_{2^{l-1}})=s_1+\sum _{j=1}^l(s_{2^j}-s_{2^{j-1}}).$$ Now we take a closer look to the number $s_{2^j}-s_{2^{j-1}}$: By definition of $s_n$, we have $$s_{2^j}-s_{2^{j-1}}=\sum _{k=2^{j-1}+1}^{2^j}\frac{1}{k}>\sum _{k=2^{j-1}+1}^{2^j}\frac{1}{2^j}=2^{j-1}\frac{1}{2^j}=\frac{1}{2}.$$ The inequality in the above formula holds true since every summand is replaced by the smallest summand $\frac{1}{2^j}$. The second last equality sign then comes from the fact that the number $\frac{1}{2^j}$ is summed up $2^{j-1}$-times. Now using this inequality together with the above sum representation for $s_{2^l}$, we obtain $$s_{2^l}=s_1+\sum _{j=1}^l(s_{2^j}-s_{2^{j-1}})>1+\sum _{j=1}^l\frac{1}{2}=1+\frac{l}{2}.$$ As a consequence, the subsequence $(s_{2^l}{)}_{l\in \mathbb{N}}$ is unbounded.$◻$

3. For $\alpha >1$, the sequence $$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{\alpha }}$$ is convergent.
   Proof: The sequence of partial sums is strictly monotonically increasing due to $$s_{n+1}-s_n=\frac{1}{(n+1{)}^{\alpha }}\ge 0.$$ Therefore, by theorems about convergence of sequences, the convergence of $(s_n{)}_{n\in \mathbb{N}}$ is shown if we find some bounded subsequence $(s_{n_j}{)}_{j\in \mathbb{N}}$. Again we use the representation for $s_{2^j}-s_{2^{j-1}}$ as in example b). We can estimate $$\begin{array}{rl}{s}_{2^j}-s_{2^{j-1}}=& \sum _{k=2^{j-1}+1}^{2^j}\frac{1}{k^{\alpha }}<\sum _{k=2^{j-1}+1}^{2^j}\frac{1}{(2^{j-1}+1{)}^{\alpha }}\\ =& \frac{2^{j-1}}{(2^{j-1}+1{)}^{\alpha }}<\frac{2^{j-1}}{(2^{j-1}{)}^{\alpha }}={\left(\frac{2}{2^{\alpha }}\right)}^{j-1}={\left(\frac{1}{2^{\alpha -1}}\right)}^{j-1},\end{array}$$ so we have $s_{2^j}-s_{2^{j-1}}<q^{j-1}$ for $q=\frac{1}{2^{\alpha -1}}$ and, due to $\alpha >1$, it holds that $0<q<1$. Using that $s_1=1=q^0$, we obtain $$s_{2^l}=s_1+\sum _{j=1}^l(s_{2^j}-s_{2^{j-1}})<1+\sum _{j=0}^{l-1}{q}^j=1+\frac{1-q^l}{1-q}<1+\frac{1}{1-q}.$$ Hence, the sequence $(s_{2^l}{)}_{l\in \mathbb{N}}$ is bounded. This implies the desired result.$◻$