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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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You can simplify the $\sin$: a known convergence test will appear. –  Davide Giraudo May 30 '12 at 21:28
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It is usually recommended that you give your thoughts on the matter, so that your fellow users might be directed properly when answering your inquiry. It is also considered ungentle to use the imperative. –  Pedro Tamaroff May 30 '12 at 21:29
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While Dirichlet's test is dandy, a beginning Calculus student might notice the $\sin$ terms give alternating signs, and then go on to use the Alternating Series Test, as hinted by Davide and Marvis. –  David Mitra May 30 '12 at 21:32
    
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;) –  Charles May 30 '12 at 21:40

2 Answers 2

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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The last series is meaningless. Maybe you should be talking about the partial sums. –  Pedro Tamaroff May 30 '12 at 21:38
    
Of course, that's the intention. Although I wouldn't call that series "meaningless": divergent, yes. Thanx. –  DonAntonio May 30 '12 at 21:54

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