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Are the following series convergent? If yes, where will they converge?

$$\sum_{n=1}^\infty{\frac{\sin{e^n}} {n}}$$

$$\sum_{n=1}^\infty\left({\frac{\sin{e^n}} {n}}\right)^2$$

$$\sum_{n=1}^\infty{\frac{\arctan{e^n}} {n}}$$

$$\sum_{n=1}^\infty{\frac{\cos{e^n}} {n}}$$

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    what do you mean where? this series are either divergent or spit out a real number...2011-03-27
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    @yoyo thanks. where means to what value they converge if they2011-03-27

3 Answers 3

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I don't think 1) and 4) are known. They depend on how close $2 e^n/\pi$ is to even and odd integers. These are hard questions. 2) and 3) are easy, though.

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    For 1) and 4), would one be prone to guess that they converge because the numerator will be random in [-1,1]? Of course, that is no help in determining the limit.2011-03-27
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    WolframAlpha says that (1) converges to about 1.19261 and that (2) converges to -1.27659.2011-03-27
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    @Zach Langley:WolframAlpha most probably uses numerical evaluation of the sum with some sort of extrapolation (which "improves" convergence). There is no way you can determine numerically whether such a sum converges or diverges.2011-03-27
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    For an interesting one along these lines: show that $\sum_{n=0}^\infty \sin((\pi/2) \phi^n)/n$ diverges, where $\phi = (1+\sqrt{5})/2$.2011-03-27
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1) and 4) are (basically) the same and i dont know...might need to bound partial sums $\sum\sin e^n$ to use some of the basic convergence tests. for 2) compare with $\sum 1/n^2$. for 3) $\arctan x\to\pi/2$ as $x\to\infty$, so compare to $\sum 1/n$.

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Maybe you can try to apply Dirichlet's test, but you need to bound the sums $\sum_{n=1}^N \sin e^n$.