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

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SUMMARY

The sequence nsin(2πen!) converges based on the analysis of its components. By expressing e as a Taylor series, specifically e = 1 + 1 + 1/2! + 1/3! + ..., and multiplying by n!, a pattern emerges that aids in determining the convergence. The presence of multiples of 2π within the sine function suggests periodic behavior, while the factorial growth of n! influences the overall convergence characteristics. Thus, the sequence exhibits predictable convergence behavior as n approaches infinity.

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Homework Statement


Does [itex]nsin(2\pi en!)[/itex]
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.
[itex]nsin(2\pi (1+1+\frac{1}{2!}+\frac{1}{3!}...)n!)[/itex]
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 [itex]nsin(2\pi en!)[/itex]
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.
[itex]nsin(2\pi (1+1+\frac{1}{2!}+\frac{1}{3!}...)n!)[/itex]
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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