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Do a convergence tests for the following series.

$\sum_{n=1}^{\infty} \frac{\log^{100}n}{n}\sin\frac{(2n+1)\pi}{2}$

I have one more question how can I show convergence with alternating series test (Leibniz's test)?

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    Peter - you've right. I'm sorry but i'm not fluent in english especially math equation and i can't share my ideas how i can do this task. Thanks for answears, now i start analyze this and i hope so i understand this;)2012-05-30

3 Answers 3

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This is an application of the Dirichlet's test (in fact the usual alternating test will do fine here) also termed the generalized alternating test. The test goes as follows. If we have a sequence of positive real numbers, $\{a_n\}_{n=1}^{\infty}$ and a sequence of complex numbers $\{b_n\}_{n=1}^{\infty}$ such that

  1. $a_n$ is a decreasing sequence i.e. $a_n > a_m$ whenever $m>n$.
  2. $\lim_{n \rightarrow \infty} a_n = 0$.
  3. $\left \lvert \displaystyle \sum_{k=0}^{n} b_k\right \rvert \leq M$, $\forall n \in \mathbb{Z}^+$, where $M$ is some constant independent of $n$

then the series $\displaystyle\sum_{n=0}^{\infty} a_n b_n$ converges.

Here let $a_n = \dfrac{\log^{100}n}{n}$ and $b_n = \sin \left(\dfrac{(2n+1) \pi}{2} \right)$.

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Apply Dirichlet's test: the seq. $\,\left\{\frac{\log^{100}n}{n}\right\}\,$ is monot. convergent to zero (after some adequate index and on), and the sequence of partial sums of the series $\sum_{n=1}^\infty\sin\frac{2n+1}{2}\pi$ is clearly bounded, thus the original series converges.

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    Of course, that's the intention. Although I wouldn't call that series "meaningless": divergent, yes. Thanx.2012-05-30