40A05-DirichletsConvergenceTest.tex

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\pmtitle{Dirichlet's convergence test}
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\pmtype{Theorem}
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\begin{document}
Theorem. Let $\{a_n\}$ and $\{b_n\}$ be sequences of real numbers such that $\{\sum_{i=0}^n a_i\}$ is bounded and $\{b_n\}$ decreases with $0$ as limit.
Then $\sum_{n=0}^\infty a_nb_n$ converges.

Proof. Let $A_n:=\sum_{i=0}^n a_n$ and let $M$ be an upper bound for $\{|A_n|\}$. By Abel's lemma, 

\begin{eqnarray*}
\sum_{i=m}^n a_ib_i &=& \sum_{i=0}^n a_ib_i - \sum_{i=0}^{m-1} a_ib_i\\
                    &=& \sum_{i=0}^{n-1} A_i(b_i-b_{i+1}) - \sum_{i=0}^{m-2} A_i(b_i-b_{i+1}) +A_nb_n - A_{m-1}b_{m-1}\\
&=&\sum_{i=m-1}^{n-1}A_i(b_i-b_{i+1}) +A_nb_n -A_{m-1}b_{m-1}\\
|\sum_{i=m}^{n} a_ib_i|&\leq& \sum_{i=m-1}^{n-1}|A_i(b_i-b_{i+1})| + |A_nb_n| + |A_{m-1}b_{m-1}|\\ 
&\leq& M \sum_{i=m-1}^{n-1}(b_i-b_{i+1}) + |A_nb_n| + |A_{m-1}b_{m-1}|\\
\end{eqnarray*}

Since $\{b_n\}$ converges to $0$, there is an $N(\epsilon)$ such that both $\sum_{i=m-1}^{n-1}(b_i-b_{i+1})<\frac{\epsilon}{3M}$ and $b_i<\frac{\epsilon}{3M}$ for $m,n>N(\epsilon)$. Then, for $m,n>N(\epsilon)$, $|\sum_{i=m}^n a_ib_i|<\epsilon$ and $\sum a_nb_n$ converges.
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