Does nsin(2πen!) have a predictable pattern for convergence?

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Homework Statement


Does nsin(2\pi en!)
converge and if so what does it converge to.
this is a sequence and n is a positive integer.

The Attempt at a Solution


My teacher gave us a hint and to write e in a Taylor series.
nsin(2\pi (1+1+\frac{1}{2!}+\frac{1}{3!}...)n!)
so we could multiply the n! through to the e stuff. And then look for a pattern their to see what it converges to.
 
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cragar said:

Homework Statement


Does nsin(2\pi en!)
converge and if so what does it converge to.
this is a sequence and n is a positive integer.

The Attempt at a Solution


My teacher gave us a hint and to write e in a Taylor series.
nsin(2\pi (1+1+\frac{1}{2!}+\frac{1}{3!}...)n!)
so we could multiply the n! through to the e stuff. And then look for a pattern their to see what it converges to.

Well, what are your thoughts? I see a bunch of stuff that is a multiple of 2pi in the sin function. But I also see a bunch of stuff that's not.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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