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Proving that the sequence $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$ is boundedly convergent on $\mathbb{R}$
Is the sum of sin(n)/n convergent or divergent?

Give a demonstration that the following series converge:


In the demonstration we can use only elementary convergence test, for example the Leibniz's test, condensation test, absolute convergence, ecc... ( these test are known by every student of the first course of analysis 1 )


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There is a full solution here at this thread: math.stackexchange.com/questions/13490/…. –  Haskell Curry Jan 4 '13 at 2:20
Does every student of the first course of analysis know how to express sines in terms of exponentials? and do such students know about the Maclaurin series for the logarithm? –  Gerry Myerson Jan 4 '13 at 2:20
See this post also. –  David Mitra Jan 4 '13 at 2:22
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marked as duplicate by David Mitra, Marvis, MJD, Alexander Gruber, Haskell Curry Jan 4 '13 at 3:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1 Answer

You can conclude it based on Abel partial summation (The result is termed as generalized alternating test or Dirichlet test). We will prove the generalized statement first.

Consider the sum $S_N = \displaystyle \sum_{n=1}^N a(n)b(n)$. Let $A(n) = \displaystyle \sum_{n=1}^N a(n)$. If $b(n) \downarrow 0$ and $A(n)$ is bounded, then the series $\displaystyle \sum_{n=1}^{\infty} a(n)b(n)$ converges.

First note that from Abel summation, we have that $$\sum_{n=1}^N a(n) b(n) = \sum_{n=1}^N b(n)(A(n)-A(n-1)) = \sum_{n=1}^{N} b(n) A(n) - \sum_{n=1}^N b(n)A(n-1)\\ = \sum_{n=1}^{N} b(n) A(n) - \sum_{n=0}^{N-1}^N b(n+1)A(n) = b(N) A(N) - b(1)A(0) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$ Now if $A(n)$ is bounded i.e. $\vert A(n) \vert \leq M$ and $b(n)$ is decreasing, then we have that $$\sum_{n=1}^{N-1} \left \vert A(n) \right \vert (b(n)-b(n+1)) \leq \sum_{n=1}^{N-1} M (b(n)-b(n+1))\\ = M (b(1) - b(N)) \leq Mb(1)$$ Hence, we have that $\displaystyle \sum_{n=1}^{N-1} \left \vert A(n) \right \vert (b(n)-b(n+1))$ converges and hence $$\displaystyle \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$ converges absolutely. Now since $$\sum_{n=1}^N a(n) b(n) = b(N) A(N) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$ we have that $\displaystyle \sum_{n=1}^N a(n)b(n)$ converges. In your case, $a(n) = \sin(n)$. Hence, $$A(N) = \displaystyle \sum_{n=1}^N a(n) = \dfrac{\sin((N+1)/2) \sin(N/2)}{\sin(1/2)} \leq \csc(1/2)$$ is bounded. Also, $b(n) = \dfrac1n$ is a monotone decreasing sequence converging to $0$.

Hence, we have that $$\sum_{n=1}^N \dfrac{\sin(n)}n$$ converges.

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I think it is time to create a post on Dirichlet test and close some infinite posts as abstract duplicates. I have answered more or less same questions over the last two days nearly $4$ times. –  user17762 Jan 4 '13 at 2:23
Is this an adaptation of an earlier solution of yours? Haha... :) –  Haskell Curry Jan 4 '13 at 2:24
@HaskellCurry Yes. Everything is the same except the last two lines where I change $a(n)$ and $b(n)$. Thats why I left the earlier comment, stating that one must create a post on Dirichlet test and close lot of these questions as abstract duplicates. –  user17762 Jan 4 '13 at 2:25
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