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Dirichlet's convergence test (Theorem)

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 $ \{\vert A_n\vert\}$. By Abel's lemma,


$\displaystyle \sum_{i=m}^n a_ib_i$ $\displaystyle =$ $\displaystyle \sum_{i=0}^n a_ib_i - \sum_{i=0}^{m-1} a_ib_i$  
  $\displaystyle =$ $\displaystyle \sum_{i=0}^{n-1} A_i(b_i-b_{i+1}) - \sum_{i=0}^{m-2} A_i(b_i-b_{i+1}) +A_nb_n - A_{m-1}b_{m-1}$  
  $\displaystyle =$ $\displaystyle \sum_{i=m-1}^{n-1}A_i(b_i-b_{i+1}) +A_nb_n -A_{m-1}b_{m-1}$  
$\displaystyle \vert\sum_{i=m}^{n} a_ib_i\vert$ $\displaystyle \leq$ $\displaystyle \sum_{i=m-1}^{n-1}\vert A_i(b_i-b_{i+1})\vert + \vert A_nb_n\vert + \vert A_{m-1}b_{m-1}\vert$  
  $\displaystyle \leq$ $\displaystyle M \sum_{i=m-1}^{n-1}(b_i-b_{i+1}) + \vert A_nb_n\vert + \vert A_{m-1}b_{m-1}\vert$  

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)$, $ \vert\sum_{i=m}^n a_ib_i\vert<\epsilon$ and $ \sum a_nb_n$ converges.



"Dirichlet's convergence test" is owned by lieven. [ full author list (2) ]
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Cross-references: Abel's lemma, upper bound, proof, converges, limit, bounded, real numbers, sequences
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This is version 2 of Dirichlet's convergence test, born on 2002-12-27, modified 2008-06-20.
Object id is 3844, canonical name is DirichletsConvergenceTest.
Accessed 8245 times total.

Classification:
AMS MSC: 40A05 (Sequences, series, summability :: Convergence and divergence of infinite limiting processes :: Convergence and divergence of series and sequences)
Pending Errata and Addenda
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questions re: Dirichlet's convergence by tangomangofandango on 2007-12-04 00:27:10
Should the second line be the summation from i=0 to i=n-1 and the summation from i=0 to i=m-2?

Also, in the last line, $ \vert\sum_{i=m}^n a_ib_i\vert<\epsilon$ < infinity, therefore $ \sum a_nb_n$ converges?

Thanks.
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