Finding the Limit of nsin(2πn!e)

[dirk_mec1]

Homework Statement

Find this limit

$$\lim_{n \rightarrow \infty} n\cdot sin(2 \pi n!e)$$

Homework Equations

hint: use the euler power series

The Attempt at a Solution

Well I know that:

$$e = \sum_{k=0}^{\infty} \frac{1}{k!}$$

But how can I use this information?

 

[G01]

Homework Statement

Find this limit

$$\lim_{n \rightarrow \infty} n\cdot sin(2 \pi n!e)$$

Homework Equations

hint: use the euler power series

The Attempt at a Solution

Well I know that:

$$e = \sum_{k=0}^{\infty} \frac{1}{k!}$$

But how can I use this information?

That is the power series representation for e.

What is the power series representation of the sine function?

 

[dirk_mec1]

$$\sin(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}+\cdots$$

Do I have to put one series into the other?

 

[m_s_a]

Unfortunately, I have erred in solution
Equal second incorrect

 

[dirk_mec1]

Unfortunately, I have erred in solution
Equal second incorrect

What do you mean?

 

[Dick]

Look at the terms in the argument of the sine for a large value of n after you've used the power series expansion for e. Many are integer multiples of 2pi. Now look at the first one that isn't. Can you justify ignoring the terms after that? Think l'Hopital.

 

[G01]

Look at the terms in the argument of the sine for a large value of n after you've used the power series expansion for e. Many are integer multiples of 2pi. Now look at the first one that isn't. Can you justify ignoring the terms after that? Think l'Hopital.

Good call Dick. I don't know what I was trying to do when I posted before about the sine expansion. That's what I get for posting at work...

 

[vipulsilwal]

i think answer is 0

n!*e comes out to be nP0+nP1+nP2+nP3....all integers
sin(2pi*integer)==0
infinity * 0=0

 

[dirk_mec1]

i think answer is 0

n!*e comes out to be nP0+nP1+nP2+nP3....all integers
sin(2pi*integer)==0
infinity * 0=0

You're wrong the answer isn't zero.

 

[vipulsilwal]

would you please tell where i am going wrong...

 

[Dick]

would you please tell where i am going wrong...

Put in the infinite series for e. A lot of terms are multiples of two pi. They aren't all. The first one that isn't contributes 2*pi/(n+1) to the sine argument. (There are more but they can be ignored relative to this one for large n).

 

[vipulsilwal]

Put in the infinite series for e. A lot of terms are multiples of two pi. They aren't all. The first one that isn't contributes 2*pi/(n+1) to the sine argument. (There are more but they can be ignored relative to this one for large n).

how n+1??
both the n(one of the n! and other of the e expansion) go upto infinity..

so it would expand to:
2pi + 2pi(n) + 2pi(n)(n-1)+ 2pi(n)(n-1)(n-2) +......2pi(n!)
2pi(n!) being the largest one.

please, more explanation needed ..

 

[Dick]

Insert the power series first. Then let n go to infinity. Don't try and do both at the same time. The 2*pi*n! term comes from the k=0 term in the expansion, 2*pi comes from the k=n term, right? What about the k=n+1 term?

 

[vipulsilwal]

i am not getting it.
could you please send me the whole solution..

 

[Dick]

i am not getting it.
could you please send me the whole solution..

We aren't in the business of providing 'whole solutions'. If you have another question ask it. I don't see what you aren't 'getting'. Put infinite series in sine argument - decide what terms are important. Take limit.