Evaluating Limits: Tools for Dealing with Complex Series

[estro]

By what tools can we evaluate limits like this:
Lim (sin((n+1)/n^2) + sin((n+2)/n^2) + ... + sin((2n)/n^2)) as n->infinity

I've played around with Maclaurin polynomials, squeeze and Stolz–Cesàro theorems...

 

[ƒ(x)]

I would say that the limit is 0 and would reach this conclusion by looking at the value of each term as n --> ∞. I don't know if there's a specific method for this kind of limit.

 

[Dick]

I would say that the limit is 0 and would reach this conclusion by looking at the value of each term as n --> ∞. I don't know if there's a specific method for this kind of limit.

Each term goes to zero but the number of terms goes to infinity, so that's not good enough. On the other hand this problem doesn't look all that subtle. Just estimate an upper bound for the size of the nth sum.

 

[estro]

the best estimate I was able to get is 1<=limit<=2

 

[Landau]

Well, that is not so bad, since the limit equals 3/2 :)

Using \sin x = x + \mathcal{O}(x^3), we get

\sum_{i=0}^n\sin\left(\frac{n+i}{n^2}\right)=\sum_{i=0}^n \frac{n+i}{n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)=\frac{3(n+1)}{2n}+\mathcal{O}\left(\frac{1}{n^3}\right)\to\frac{3}{2}.

 

[estro]

Well, that is not so bad, since the limit equals 3/2 :)

Using \sin x = x + \mathcal{O}(x^3), we get

\sum_{i=0}^n\sin\left(\frac{n+i}{n^2}\right)=\sum_{i=0}^n \frac{n+i}{n^2}.

I can't understand why this true?

 

[Cyosis]

I can't understand why this true?

It is not true, but that's not what he wrote. Don't forget that big O added to that term. What he has done is expand the sine using its power series. If you do that you will notice that the first term is (n+i)/n^2 while the rest is of order n^-3 or smaller.

 

[estro]

Thanks!

 

[Landau]

You should quote correctly ;)

Using \sin x = x + \mathcal{O}(x^3), we get

\sum_{i=0}^n\sin\left(\frac{n+i}{n^2}\right)=\sum_{i=0}^n \frac{n+i}{n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)

 

[estro]

You should quote correctly ;)

I still have hard time understanding this =( (I'm new to taylor series)
By the way I trying to calculate series limit...

 

[Landau]

Well, tell us what you don't understand? Do you understand sin x = x + O(x^3), i.e. do you know the taylor series of the sine?

 

[estro]

Well, tell us what you don't understand? Do you understand sin x = x + O(x^3), i.e. do you know the taylor series of the sine?

I Don't understand these points:
1. \sum_ {i=0}^n \frac{n+i}{n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)=\frac{3(n+1)}{2n}+\mathcal{O}\left(\frac{1}{n^ 3}\right)
2. How can we approximate series using Taylor polynomials? (they don't have derivative)

Thanks!

 

[Landau]

The only approximation that is made, is sin x = x +O(x^3). The rest is just substituting in the sum. I used that

\sum_ {i=0}^n \frac{n+i}{n^2}=\frac{3(n+1)}{2n}.

This is easily calculated if you know that

\sum_ {i=0}^n i=\frac{n(n+1)}{2}.

 

[estro]

Oh thanks,
but can we use this directly on a series?