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

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In summary, the given problem involves determining whether the sequence nsin(2pi en!) converges and what it converges to. The suggested approach is to write e in a Taylor series and then multiply it by n! to look for a pattern. The sequence involves multiples of 2pi in the sin function, but also includes other terms.
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cragar
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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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FAQ: Does nsin(2πen!) have a predictable pattern for convergence?

What is the formula for nsin(2πen!)?

The formula for nsin(2πen!) is n times the sine of 2π multiplied by the factorial of n.

How do you determine if nsin(2πen!) converges?

To determine if nsin(2πen!) converges, we use the ratio test or the root test. If the limit of the ratio or root is less than 1, then the series converges. If it is greater than 1, then the series diverges.

What is the significance of the factorial in nsin(2πen!)?

The factorial in nsin(2πen!) represents the number of terms in the series. As n increases, the factorial also increases, resulting in more terms and potentially affecting the convergence or divergence of the series.

Does the value of e affect the convergence of nsin(2πen!)?

Yes, the value of e does affect the convergence of nsin(2πen!). As e increases, the terms in the series also increase, potentially affecting the convergence or divergence of the series.

Can nsin(2πen!) converge to a specific value?

No, nsin(2πen!) does not converge to a specific value. It either converges to a finite value or diverges. The value of n can affect the convergence to a certain extent, but the series will not converge to a specific value.

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