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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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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.

There is a full solution here at this thread:…. –  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

1 Answer 1

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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