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We know that the Dirichlet criterion testing the convergence of a series needs one sequence monotonically tends to zero, and the sequence of partial sums of the the other series is bounded.

I wonder if there is a counterexample of it if the first sequence is not monotone?

Much thanks to your help~

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    A better statement of the Dirichlet Test is that $\left|\sum a_n\right|\le A$, $b_n\to0$, and $\sum\left|b_{n+1}-b_n\right|\le B$, then $\sum a_nb_n$ converges to a limit whose absolute value is less than $AB$. This works for complex sequences, too.2013-03-13

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For $n\ge 2$, let $a_n=\dfrac{(-1)^n}{\log n}$. Let $b_n=\dfrac{1}{\log n}$ if $n$ is even, and let $b_n=\dfrac{1}{2^n}$ if $n$ is odd. Then $\sum_2^\infty a_nb_n$ diverges. This is because the sum of the negative terms converges rapidly, while the sum of the positive terms is unbounded.

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    Andre Nicolas, I still don't get how does that answer the question: Are there any counterexamples to the Dirichlet criterion if g(x) is not monotone?2013-03-13
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This shows the monotone condition is required in the Leibniz alternating series test, of which the Dirichlet test is a generalization of.

Consider the series $\displaystyle \sum_{n=2}^{\infty} a_n$ where $\displaystyle a_{2k} = \frac{1}{k}$ and $\displaystyle a_{2k+1} = -\frac{1}{2k} $. Then $ \sum_{n=2}^{2m} a_n = \left( \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{m} \right) - \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2(m-1)} \right) = \frac{1}{2} \sum_{k=1}^{m-1} \frac{1}{k} + \frac{1}{m} $

which is almost the Harmonic series, and thus divergent.

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    Lol. "Almost the Harmonic series" sounds $f$unny. +12012-04-25