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

1.$\pi/3$

2.$\pi/4$

3.$\pi/2$

4.None

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. $\pi/e$

Gunjan
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    I guess you expect someone to edit your question and make it readable. – Julien 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
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    @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: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Dennis Gulko Mar 10 '13 at 17:04
  • Shouldn't that 18 in the first line be a 16? Otherwise I don't understand how that series is formed. – DonAntonio Mar 10 '13 at 17:23
  • No, its 18..Its the question from an entrance exam..Can't say if its wrong.. – Gunjan Mar 10 '13 at 17:31
  • In case its 16 instead of 18 then how to solve this type of series – Gunjan Mar 10 '13 at 17:38
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    @DonAntonio If it's 18 then the sequence is $2n^2$. – MJD Mar 10 '13 at 18:48
  • Thanks, @MJD .Only now that's clear to me... – DonAntonio Mar 10 '13 at 20:37

3 Answers3

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$(2)$ 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.

lab bhattacharjee
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    @user1709828, I'm trying to identify whether there is any general formula to determine the telescopic sum form. A similar problem : http://math.stackexchange.com/questions/193001/explicitly-finding-the-sum-of-arctan1-n2n1 – lab bhattacharjee Mar 10 '13 at 18:42
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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 $0$

Ross Millikan
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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.