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Jiří Lebl, Basic Analysis I–II, version 6.3. Free author edition of this section. Selection and attribution · Notation.

L2.6.5: Full written Mertens proof; real comparisons only on moduli, with the complex extension stated explicitly in P13.2.

Proof.

Suppose ∑n=0∞an\sum_{n=0}^\infty a_n converges absolutely, and let ϵ>0\epsilon > 0 be given. In this proof, instead of picking complicated estimates just to make the final estimate come out as less than ϵ,\epsilon\text{,} let us simply obtain an estimate that depends on ϵ\epsilon and can be made arbitrarily small.
Write
Am≔∑n=0man,Bm≔∑n=0mbn.\begin{equation*} A_m \coloneqq \sum_{n=0}^m a_n , \qquad B_m \coloneqq \sum_{n=0}^m b_n . \end{equation*}
We rearrange the mmth partial sum of ∑n=0∞cn:\sum_{n=0}^\infty c_n\text{:}
∣(∑n=0mcn)−AB∣=∣(∑n=0m∑i=0naibn−i)−AB∣=∣(∑n=0mBnam−n)−AB∣=∣(∑n=0m(Bn−B)am−n)+BAm−AB∣≤(∑n=0m∣Bn−B∣∣am−n∣)+∣B∣∣Am−A∣\begin{equation*} \begin{split} \abs{\left(\sum_{n=0}^m c_n \right) - AB} & = \abs{\left( \sum_{n=0}^m \sum_{i=0}^n a_i b_{n-i} \right) - AB} \\ & = \abs{\left( \sum_{n=0}^m B_n a_{m-n} \right) - AB} \\ & = \abs{\left( \sum_{n=0}^m ( B_n - B ) a_{m-n} \right) + B A_m - AB} \\ & \leq \left( \sum_{n=0}^m \sabs{ B_n - B } \sabs{a_{m-n}} \right) + \sabs{B}\sabs{A_m - A} \end{split} \end{equation*}
We can surely make the second term on the right-hand side small. The trick is to handle the first term. Pick KK such that for all m≥K,m \geq K\text{,} we have ∣Am−A∣<ϵ\sabs{A_m - A} < \epsilon and also ∣Bm−B∣<ϵ.\sabs{B_m - B} < \epsilon\text{.} As ∑n=0∞an\sum_{n=0}^\infty a_n converges absolutely, there is a KK large enough such that for all m≥K,m \geq K\text{,}
∑n=Km∣an∣<ϵ.\begin{equation*} \sum_{n=K}^m \sabs{a_n} < \epsilon . \end{equation*}
As ∑n=0∞bn\sum_{n=0}^\infty b_n converges, Bmax≔sup⁡{∣Bn−B∣:n=0,1,2,…}B_{\text{max}} \coloneqq \sup \bigl\{ \sabs{ B_n - B } : n = 0,1,2,\ldots \bigr\} is finite. Take m≥2K.m \geq 2K\text{.} In particular, m−K+1>K.m-K+1 > K\text{.} So
∑n=0m∣Bn−B∣∣am−n∣=(∑n=0m−K∣Bn−B∣∣am−n∣)+(∑n=m−K+1m∣Bn−B∣∣am−n∣)≤(∑n=Km∣an∣)Bmax+(∑n=0K−1ϵ∣an∣)≤ϵBmax+ϵ(∑n=0∞∣an∣).\begin{equation*} \begin{split} \sum_{n=0}^m \sabs{ B_n - B } \sabs{a_{m-n}} & = \left( \sum_{n=0}^{m-K} \sabs{ B_n - B } \sabs{a_{m-n}} \right) + \left( \sum_{n=m-K+1}^m \sabs{ B_n - B } \sabs{a_{m-n}} \right) \\ & \leq \left( \sum_{n=K}^m \sabs{a_{n}} \right) B_{\text{max}} + \left( \sum_{n=0}^{K-1} \epsilon \sabs{a_{n}} \right) \\ & \leq \epsilon B_{\text{max}} + \epsilon \left( \sum_{n=0}^\infty \sabs{a_{n}} \right) . \end{split} \end{equation*}
Therefore, for m≥2K,m \geq 2K\text{,} we have
∣(∑n=0mcn)−AB∣≤(∑n=0m∣Bn−B∣∣am−n∣)+∣B∣∣Am−A∣≤ϵBmax+ϵ(∑n=0∞∣an∣)+∣B∣ϵ=ϵ(Bmax+(∑n=0∞∣an∣)+∣B∣).\begin{equation*} \begin{split} \abs{\left(\sum_{n=0}^m c_n \right) - AB} & \leq \left( \sum_{n=0}^m \sabs{ B_n - B } \sabs{a_{m-n}} \right) + \sabs{B}\sabs{A_m - A} \\ & \leq \epsilon B_{\text{max}} + \epsilon \left( \sum_{n=0}^\infty \sabs{a_{n}} \right) + \sabs{B}\epsilon = \epsilon \left( B_{\text{max}} + \left( \sum_{n=0}^\infty \sabs{a_{n}} \right) + \sabs{B} \right) . \end{split} \end{equation*}
The expression in the parentheses on the right-hand side is a fixed number. Hence, we can make the right-hand side arbitrarily small by picking a small enough ϵ>0.\epsilon> 0\text{.} So ∑n=0∞cn\sum_{n=0}^\infty c_n converges to AB.AB\text{.}