# $\cos(x)$ and $\arccos(x)$ couple limit

Find the value of the following limit:

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

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Are you limiting $\frac{cos(1)arccos(\frac{1}{n})+cos(\frac{1}{2})arccos(\frac{1}{n-1})+...+cos(\‌​frac{1}{n})arccos(1)}{n}$ ? –  Nancy Rutkowskie May 25 '12 at 17:49
@Nancy R: i posted the limit above. Yes. –  Chris's sis the artist May 25 '12 at 17:54
@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. –  data May 25 '12 at 17:59
@m.woj: thanks for your comment –  Chris's sis the artist May 25 '12 at 18:06

## 3 Answers

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

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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. –  Chris's sis the artist May 25 '12 at 19:12
thanks for your solution. It's a bit ugly but i need to get used with these not-so-nice approaches. –  Chris's sis the artist May 25 '12 at 20:17

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.

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+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. –  Sasha May 25 '12 at 19:42
This is becoming a community proof now. Thanks @Sasha! –  Egbert May 25 '12 at 20:34
@Egbert: thanks for your interesting proof. –  Chris's sis the artist May 25 '12 at 20:48
Great proof Egbert! I'm trying to come up with one involving integration... –  Pedro Tamaroff May 25 '12 at 20:50
...but it seems fruitless. –  Pedro Tamaroff May 25 '12 at 20:54

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.

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@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. –  Dejan Govc May 26 '12 at 17:52
@Chris: I just noticed the rearrangement inequality didn't really add anything. How silly of me. (I have edited the answer accordingly.) –  Dejan Govc May 27 '12 at 12:15