11
$\begingroup$

Find the value of the following limit:

$$\lim_{n\to\infty} \frac {\cos 1 \cdot \arccos \frac{1}{n}+\cos\frac {1}{2} \cdot \arccos \frac{1}{(n-1)}+ \cdots +\cos \frac{1}{n} \cdot \arccos{1}}{n}$$

  • 0
    Are you limiting $\frac{cos(1)arccos(\frac{1}{n})+cos(\frac{1}{2})arccos(\frac{1}{n-1})+...+cos(\frac{1}{n})arccos(1)}{n}$ ?2012-05-25
  • 0
    @Nancy R: i posted the limit above. Yes.2012-05-25
  • 0
    @Chris My first though was to treat the sum in the numerator as a series and try to sum it up to, say - $k$ then bound given sequence from above by the sequence $\frac{k}{n}$ and from below by constant function $0$. But I can't see immediatly if it's the shortest way to evaluate that limit or if some extra difficulties wouldn't appear.2012-05-25
  • 0
    @m.woj: thanks for your comment2012-05-25

3 Answers 3

4

This is too long for a comment, so I am posting it as an answer, although it doesn't completely resolve the question.

I shall prove that $\frac{\pi}{2}$ is an upper bound.

Lemma. For all $n\in\Bbb N$, we have $\frac{1}{n} \sum\limits_{k=1}^n \cos\left(\frac{1}{k}\right) \arccos\left( \frac{1}{n+1-k} \right) \leq \frac{\pi}{2}$.

Proof. Since both functions are decreasing, we have: $$\frac{1}{n} \sum_{k=1}^n \cos\left(\frac{1}{k}\right) \arccos\left( \frac{1}{n+1-k} \right) \leq \frac1nn\cos(0)\arccos(0)=\frac{\pi}{2},$$

which shows the desired inequality. $\square$

As Sasha mentions above this is probably also the value of the limit.

As Chris notes in the comment below this answer, it is possible to prove that $\frac{\pi}2$ is also a lower bound by a simple application of AM-GM inequality and the Cesaro-Stolz theorem, completing the proof.

  • 0
    @Chris: That's a great idea! There may be some complications caused by the fact that $\arccos1=0$ but we can simply throw that last term away and write $\frac1n=\frac1{n-1}\frac{n-1}n$, allowing us to use AM-GM on the remaining $n-1$ terms, which are positive.2012-05-26
  • 0
    @Chris: I just noticed the rearrangement inequality didn't really add anything. How silly of me. (I have edited the answer accordingly.)2012-05-27
7

A more convenient way to state the sequence is: $$ \frac{\sum_{k=1}^n\cos\frac{1}{k}\arccos\frac{1}{n-k+1}}{n} $$

Note that for $0\leq x\leq 1$ we have $1-x\leq\cos x\leq 1$ and $\frac{\pi}{2}-\frac{\pi}{2}x\leq\arccos x\leq \frac{\pi}{2}-x$. Therefore, we have $$ \tfrac{\pi}{2}(1-\tfrac{1}{k})(1-\tfrac{1}{n-k+1})\leq\cos\frac{1}{k}\arccos\frac{1}{n-k+1}\leq\frac{\pi}{2}-\frac{1}{n-k+1}. $$ Thus, a lower bound for the limit is given by the sequence $$ \frac{\sum_{k=1}^n\frac{\pi}{2}(1-\tfrac{1}{k})(1-\tfrac{1}{n-k+1})}{n}=\frac{\pi}{2}-\frac{\sum_{k=0}^n\frac{(n-k+1)+k-1}{k(n-k+1)}}{n}=\frac{\pi}{2}-\sum_{k=1}^n\frac{1}{k(n-k+1)} $$ The terms in the latter sum can be rewritten to $$ \frac{1}{k(n+1)}+\frac{1}{(n-k+1)(n+1)} $$ so we get the sums $$ \frac{1}{n+1}\sum_{k=1}^n\frac{1}{k}\qquad\text{and}\qquad\frac{1}{n+1}\sum_{k=1}^n\frac{1}{n-k+1}. $$ Sacha indicated a proof in the comments that these sums tend to zero. Similarly, for the upper bound we have $$ \frac{\sum_{k=1}^n\frac{\pi}{2}-\tfrac{1}{n-k+1}}{n}=\frac{\pi}{2}-\frac{\sum_{k=1}^n\frac{1}{n-k+1}}{n}\to\frac{\pi}{2} $$ This gives an upper bound of $\frac{\pi}{2}$. This finishes the proof that the limit is $\frac{\pi}{2}$.


Note: Another proof that $\frac{1}{n}\sum_{k=1}^n\frac{1}{k}\to 0$. Using Cauchy-Schwarz, we see that $$ \sum_{k=1}^n\frac{1}{k}\leq \sqrt{n}\sqrt{\textstyle\sum_{k=1}^n\tfrac{1}{k^2}} $$ Therefore we get $$ \frac{1}{n}\sum_{k=1}^n\frac{1}{k}\leq\sqrt{\frac{\textstyle\sum_{k=1}^n\tfrac{1}{k^2}}{n}} $$ In the square root on the right hand side, the numerator converges, so the whole tends to zero.

  • 0
    +1 This is exactly the way to go. You can now show that $\lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{n+1-k} = \lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{k} = \lim_{n\to \infty} \frac{1}{n} H_n = 0$, and similarly for the sum of the lower bound.2012-05-25
  • 0
    This is becoming a community proof now. Thanks @Sasha!2012-05-25
  • 0
    @Egbert: thanks for your interesting proof.2012-05-25
  • 0
    Great proof Egbert! I'm trying to come up with one involving integration...2012-05-25
  • 0
    ...but it seems fruitless.2012-05-25
  • 0
    @Peter: What do you mean with that?2012-05-25
  • 0
    I don't think I will get anywhere.2012-05-25
  • 0
    @Peter: could you please be more precise and say what is wrong?2012-05-25
  • 0
    I'm just saying I'm trying to solve the problem using integration but I don't think I will get anywhere. Your proof is flawless and very inspiring!2012-05-25
  • 0
    @Peter: Ah! I missed your first comment :) Sorry for that. Thanks for your kind words! (I already found it strange that you said you couldn't get anywhere... I'm glad the problem is now resolved. It wasn't automatic for me... :)2012-05-25
5

What follows is a little hand-wavy, and I wish I had more rigorous demonstration, but the post is too big for a comment.

$$ \begin{eqnarray} \frac{1}{n} \sum_{k=1}^n \cos\left(\frac{1}{k}\right) \arccos\left( \frac{1}{n+1-k} \right) &=& \frac{1}{n} \sum_{k=1}^n \left( 1 - 2 \sin^2\left(\frac{1}{2 k}\right) \right) \left( \frac{\pi}{2} - \arcsin\left( \frac{1}{n+1-k} \right) \right) \end{eqnarray} $$ We can now split the sum in two parts, $1 \leqslant k \leqslant \lfloor\frac{n}{2}\rfloor$ and $\lfloor\frac{n}{2}\rfloor < k \leqslant n$. In each of these parts, either $\sin$, or $\arcsin$ will be small, and in the limiting value will be $\frac{\pi}{2}$:

In[28]:= Table[   N[1/n Sum[Cos[1/k] ArcCos[1/(n - k + 1)], {k, 1, n}],     50], {n, {5000, 10000, 100000}}] // N  Out[28]= {1.56861, 1.56963, 1.57066} 


More rigorously: $$ \frac{1}{n} \sum_{k=1}^n \left( 1 - 2 \sin^2\left(\frac{1}{2 k}\right) \right) \left( \frac{\pi}{2} - \arcsin\left( \frac{1}{n+1-k} \right) \right) = \frac{\pi}{2} - \frac{\pi}{n} \sum_{k=1}^n \sin^2 \frac{1}{k} - \frac{1}{n} \sum_{k=1}^n \arcsin\frac{1}{k} + \frac{2}{n} \sum_{k=1}^n \sin^2\left( \frac{1}{k}\right) \arcsin\left(\frac{1}{n+1-k}\right) $$ Now: $$ 0 \leqslant \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \sin^2\left(\frac{1}{k}\right) \leqslant \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{k^2} = 0 $$ $$ 0 \leqslant \lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \arcsin\left(\frac{1}{k}\right) \leqslant \lim_{n \to \infty} \frac{\pi}{2n} \sum_{k=1}^n \frac{1}{k} = \lim_{n \to \infty} \frac{\pi}{2 n} \ln(n) = 0 $$ $$ 0 \leqslant \lim_{n \to \infty} \frac{2}{n} \sum_{k=1}^n \sin^2 \left( \frac{1}{k} \right) \arcsin\left(\frac{1}{n+1-k} \right) \leqslant \lim_{n \to \infty} \frac{2}{n} \sum_{k=1}^n \frac{1}{k^2} \frac{1}{n+1-k} \leqslant \lim_{n \to \infty} \frac{2}{n} \sum_{k=1}^n \frac{1}{k^2} = 0 $$

  • 0
    could you provide with some more details pls about that splitting. I'm surrounded by mist because maybe it's something i don't catch it yet. This is from my highschool notebook and i was thinking that there is an easier way to solve it. I'm struggling to understand your way, now.2012-05-25
  • 0
    thanks for your solution. It's a bit ugly but i need to get used with these not-so-nice approaches.2012-05-25