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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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    @m.woj: thanks for your comment2012-05-25

3 Answers 3

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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: I just noticed the rearrangement inequality didn't really add anything. How silly of me. (I have edited the answer accordingly.)2012-05-27
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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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    @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
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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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    thanks for your solution. It's a bit ugly but i need to get used with these not-so-nice approaches.2012-05-25