The limit of the expression lim_{n → ∞} n·sin(2πn!e) can be evaluated using the power series expansion of the exponential function and the sine function. The key insight is that many terms in the argument of the sine function become integer multiples of 2π, which results in sin(2π·integer) = 0. However, the first term that is not an integer multiple contributes significantly, leading to the conclusion that the limit evaluates to 0.
e and sin(x)e and sin(x) in detailStudents studying calculus, particularly those focusing on limits and series expansions, as well as educators looking for examples of limit evaluation techniques.
dirk_mec1 said: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?
m_s_a said:Unfortunately, I have erred in solution
Equal second incorrect
Dick said: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.
vipulsilwal said:i think answer is 0
n!*e comes out to be nP0+nP1+nP2+nP3....all integers
sin(2pi*integer)==0
infinity * 0=0
vipulsilwal said:would you please tell where i am going wrong...
Dick said: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 said:i am not getting it.
could you please send me the whole solution..