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Q1: The sum of the infinite series $\cot ^{-1}2 + \cot ^{-1} 8+ \cot^{-1}18+ \cot^{-1}32\cdots$





Q2: Value of $\lim_ {n \to \infty}[ {\cos \frac{\pi}{2^2} } {\cos \frac{\pi}{2^3} } \ldots{\cos \frac{\pi}{2^n} }$]

  1. $\pi$

  2. $1/\pi$

  3. $2/\pi$

  4. $\pi/e$

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I guess you expect someone to edit your question and make it readable. – 1015 Mar 10 '13 at 16:59
I don't know how to edit my question...I tried my best to write. – Gunjan Mar 10 '13 at 17:02
@user19945: Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what are your thoughts; this will prevent people from telling you things you already know, and help them give their answers at the right level. Consider editing you question. Use this$\LaTeX$ guide:… – Dennis Gulko Mar 10 '13 at 17:04
Thank u...i m trying now – Gunjan Mar 10 '13 at 17:14
@DonAntonio If it's 18 then the sequence is $2n^2$. – MJD Mar 10 '13 at 18:48

$(2)$ Applying $\sin2x=2\sin x\cos x,$ $$T_n=\prod_{2\le r\le n}\cos \frac{\pi}{2^r}$$ $$=\frac{\sin\frac{\pi}{2^{n-1}}}{2\sin\frac{\pi}{2^n}}\prod_{2\le r\le n-1}\cos \frac{\pi}{2^r}$$ $$=\frac{\sin\frac{\pi}{2^{n-2}}}{2\cdot2\sin\frac{\pi}{2^n}}\prod_{2\le r\le n-2}\cos \frac{\pi}{2^r}$$ $$=\frac{\sin\frac{\pi}{2^{n-3}}}{2^3\sin\frac{\pi}{2^n}}\prod_{2\le r\le n-3}\cos \frac{\pi}{2^r}$$


$$=\frac{\sin\frac{\pi}{2^{n-s}}}{2^s\sin\frac{\pi}{2^n}}\prod_{2\le r\le n- s}\cos \frac{\pi}{2^r}\text{ where } 0\le s\le n-2$$

Putting $s=n-2,$ $$T_n=\frac{\sin\frac{\pi}{2^{2}}}{2^{n-2}\sin\frac{\pi}{2^n}}\prod_{2\le r\le 2}\cos \frac{\pi}{2^r}=\frac{\sin\frac{\pi}{2^{2}}}{2^{n-2}\sin\frac{\pi}{2^n}}\cos\frac{\pi}{4}=\frac{2\sin\frac{\pi}{4}\cos\frac{\pi}{4}}{2^{n-1}\sin\frac{\pi}{2^n}}=\frac1{2^{n-1}\sin\frac{\pi}{2^n}}$$

Putting $2^n=\frac1y$ as $n\to\infty\implies y\to0$

So, $$\lim_{n\to\infty}\prod_{2\le r\le n}\cos \frac{\pi}{2^r}=\lim_{y\to0}\frac{2y}{\sin \pi y}=\frac2\pi\lim_{y\to0}\frac{\pi y}{\sin \pi y}=\frac2\pi$$

$(1)$ Assuming the $m$th term to be $\text{arccot}(2m^2)$

This is probably how Marvis found the Telescopic sum form of arccot$(2m^2)$

$$\text{ As arccot}x-\text{arccot}y=\text{arccot}\left(\frac{xy+1}{y-x}\right)$$ $$\text{arccot}(2m^2)=\text{arccot}\frac{m+1}m-\text{arccot}\frac m{m-1} $$

The rest is like his solution.

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(+1) for details. great. – user45099 Mar 10 '13 at 18:38
@user1709828, I'm trying to identify whether there is any general formula to determine the telescopic sum form. A similar problem :… – lab bhattacharjee Mar 10 '13 at 18:42

For 1, I don't think it is clear what the series is, so would pick 4.

For 2, $\cos \pi=-1$so the numerator is $(-1)^n$ and the denominator gets huge, so the limit is $0$

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@RosMilikan, I found the limit to be $\frac2\pi$ in my answer – lab bhattacharjee Mar 10 '13 at 17:56
Apparently, the coeffitients of 1. are $2*n^2$ (found using oeis). However, that is not definite. – CBenni Mar 10 '13 at 18:02
@CBenni, could you please have a look into my answer? – lab bhattacharjee Mar 10 '13 at 18:06
For the second, it has changed from $\frac{\cos \pi}{2^n}$ to $\cos \frac \pi{2^n}$ – Ross Millikan Mar 10 '13 at 21:18

For the first one, $$\sum_{k=1}^{m} \text{arccot}(2n^2) = \text{arccot} \left(\dfrac{m+1}m\right)$$ Your sum is $$\sum_{k=1}^{\infty} \text{arccot}(2n^2) = \lim_{m \to \infty}\text{arccot} \left(\dfrac{m+1}m\right) = \dfrac{\pi}4$$

For the second one, $$\prod_{k=1}^m \cos\left(\dfrac{\theta}{2^{k+1}} \right) = \dfrac{\sin\left(\dfrac{\theta}{2}\right)}{2^{m}\sin \left(\dfrac{\theta}{2^{m+1}} \right)}$$ Hence, your product is $$\prod_{k=1}^{\infty} \cos\left(\dfrac{\pi}{2^{k+1}} \right) = \lim_{m \to \infty} \dfrac{\sin\left(\dfrac{\pi}{2}\right)}{2^{m}\sin \left(\dfrac{\pi}{2^{m+1}} \right)} = \dfrac2{\pi}$$

For the first one, recall $$\cot(A+B) = \dfrac{\cot(A) \cot(B) - 1}{\cot(A) + \cot(B)}$$ \begin{align} \cot \left(\text{arccot}\left(\dfrac{m}{m-1}\right) + \text{arccot}\left(2m^2 \right)\right) & = \dfrac{\dfrac{m}{m-1} \cdot 2 \cdot m^2-1}{\dfrac{m}{m-1} + 2 \cdot m^2}\\ & = \dfrac{2m^3-m+1}{2m^3-2m^2+m}\\ & = \dfrac{(m+1)(2m^2-2m+1)}{m(2m^2-2m+1)}\\ & = \dfrac{m+1}m \end{align} Now use the above two along with induction to conclude what you want.

For second one, recall $\sin(2 \phi) = 2 \sin(\phi) \cos(\phi)$ and induction to conclude what you want.

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how did you obtain the very first identity? – user45099 Mar 10 '13 at 18:08
In both first and second case? – user45099 Mar 10 '13 at 18:14
@user1709828,for the second case, you may have a look into my answer. – lab bhattacharjee Mar 10 '13 at 18:16
+1 after adding the proof for the first identity. – CBenni Mar 10 '13 at 21:34

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