Find sum of the Trignomertric series 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 $\lim_ {n \to \infty}[ {\cos \frac{\pi}{2^2} } {\cos
\frac{\pi}{2^3} } \ldots{\cos \frac{\pi}{2^n} }$]


*

*$\pi$

*$1/\pi$

*$2/\pi$

*$\pi/e$
 A: $(2)$ 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.
A: 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 $0$
A: 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.
