∑ limitsr = 1∞ cot-1 (r2 + (3/4)) equals

Question

∑ limitsr = 1∞ cot-1 (r2 + (3/4)) equals (A) (π/2) (B) cot-1 2 (C) (π/6) (D) tan-1 2

Tardigrade Admin · Accepted Answer

∑ limitsr = 1∞ cot-1 (r2 + (3/4)) = ∑ limitsr=1∞ ((1/r2 + (3)4)) = ∑ limitsr=1∞ tan-1 [((r + (1/2)) - (r - (1/2))/1 + (r + (1)2)(r - (1)2)] = (tan/-1) (3/2) - tan-1 (1/2)) + (tan-1 (5/2) - tan-1 (3/2)) + (tan-1 (7/2) - tan-1 (5/2)) + ...+ (π/2) = (π/2) - tan-1 (1/2) = cot-1(1/2) = tan-1 2

Q. $r = 1 \sum \infty co t^{- 1} (r^{2} + \frac{3}{4})$ equals

$\frac{π}{2}$

$co t^{- 1} 2$

$\frac{π}{6}$

$t a n^{- 1} 2$

Q. r=1∑∞​cot−1(r2+43​) equals

2π​

cot−12

6π​

tan−12

r=1∑∞​cot−1(r2+43​) =r=1∑∞​(r2+43​1​) =r=1∑∞​tan−1[1+(r+21​)(r−21​)(r+21​)−(r−21​)​] =(tan−123​−tan−121​)+(tan−125​−tan−123​) +(tan−127​−tan−125​)+...+2π​ =2π​−tan−121​=cot−1(1/2)=tan−12

Q. $r = 1 \sum \infty co t^{- 1} (r^{2} + \frac{3}{4})$ equals

$\frac{π}{2}$

$co t^{- 1} 2$

$\frac{π}{6}$

$t a n^{- 1} 2$