Geometric and Harmonic Series

Example 1.

1. For \(q\in\mathbb{K}\), the geometric series \[\sum_{k=0}^\infty q^k\] is convergent if and only if \(|q|<1\).

   Proof: We can show that the \(n\)-th partial sum is given by \[s_n=\sum_{k=0}^nq^k=\begin{cases}\frac{1-q^{n+1}}{1-q}&:\text{ if }q\neq1,\\n+1&:\text{ if }q=1.\end{cases}\] Hence, \((s_n)_{n\in\mathbb{N}}\) is convergent if and only if \(|q|<1\). In this case we have \[\sum_{k=0}^\infty q^k=\lim_{n\to\infty}s_n=\lim_{n\to\infty}\frac{1-q^{n+1}}{1-q}=\frac{1}{1-q}.\]

2. The harmonic series \[\sum_{k=1}^\infty \frac1k\] is divergent to \(+\infty\).
   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 \[s_{2^l}=s_1+(s_2-s_1)+(s_4-s_2)+(s_8-s_4)+\ldots+(s_{2^l}-s_{2^{l-1}})=s_1+\sum_{j=1}^l(s_{2^j}-s_{2^{j-1}}).\] Now we take a closer look to the number \(s_{2^j}-s_{2^{j-1}}\): By definition of \(s_n\), we have \[s_{2^j}-s_{2^{j-1}}=\sum_{k=2^{j-1}+1}^{2^j}\frac1k>\sum_{k=2^{j-1}+1}^{2^j}\frac1{2^{j}}= 2^{j-1}\frac1{2^{j}}=\frac12.\] The inequality in the above formula holds true since every summand is replaced by the smallest summand \(\frac1{2^{j}}\). The second last equality sign then comes from the fact that the number \(\frac1{2^{j}}\) is summed up \(2^{j-1}\)-times. Now using this inequality together with the above sum representation for \(s_{2^l}\), we obtain \[s_{2^l}=s_1+\sum_{j=1}^l(s_{2^j}-s_{2^{j-1}})>1+\sum_{j=1}^l\frac12=1+\frac{l}2.\] As a consequence, the subsequence \((s_{2^l})_{l\in\mathbb{N}}\) is unbounded.\(\Box\)

3. For \(\alpha>1\), the sequence \[\sum_{k=1}^\infty \frac1{k^\alpha}\] is convergent.
   Proof: The sequence of partial sums is strictly monotonically increasing due to \[s_{n+1}-s_n=\frac1{(n+1)^\alpha}\geq0.\] 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 \[\begin{aligned} s_{2^j}-s_{2^{j-1}}=&\,\sum_{k=2^{j-1}+1}^{2^j}\frac1{k^\alpha}<\sum_{k=2^{j-1}+1}^{2^j}\frac1{(2^{j-1}+1)^\alpha}\\ =&\,\frac{2^{j-1}}{(2^{j-1}+1)^\alpha}< \frac{2^{j-1}}{(2^{j-1})^\alpha}=\left(\frac{2}{2^\alpha}\right)^{j-1}=\left(\frac{1}{2^{\alpha-1}}\right)^{j-1}, \end{aligned}\] 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.\(\Box\)