The value of Limit where n tends to ♾️ 6 tan {Summation of r=1 to n tan^-1 (1/r^2+3r+3)} is equal to

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Limit where n tends to ♾️ tan {Summation of r=1 to n tan^-1 (1/1+r+r^2)} is equal to: Limit where n tends to ♾️ tan {Summation of r=1 to n tan^-1 (1/1+r+r^2)} is equal to:
Find the value of lim_(n to oo) 6 tan{sum_(r=1)^n tan^-1(1/(r^2+3r+3))} is | CLASS 12 | JEE MAIN... Find the value of lim_(n to oo) 6 tan{sum_(r=1)^n tan^-1(1/(r^2+3r+3))} is | CLASS 12 | JEE MAIN...
lim n tends to infinity Sum r=1 to n tan^-1(2×3^r-1/1+3^2r-1) lim n tends to infinity Sum r=1 to n tan^-1(2×3^r-1/1+3^2r-1)
JEE MAINS 2021, (24th feb,1st shift) Evaluate lim n tends to infinity tan(sigma r=1 to n(tan^-1(1/. JEE MAINS 2021, (24th feb,1st shift) Evaluate lim n tends to infinity tan(sigma r=1 to n(tan^-1(1/.
Let Sk = Summation of r=1 to k tan^-1 (6^r/ 2^2r+1 + 3^2r+1). Then Lim k tends to infinity Sk = ? Let Sk = Summation of r=1 to k tan^-1 (6^r/ 2^2r+1 + 3^2r+1). Then Lim k tends to infinity Sk = ?
JEE Main 2021 Solution | SUM(t=1 to infinity) tan^-1(6^r/{2^(2r+1)+3^(2r+1)}) #JEEMAIN2021 JEE Main 2021 Solution | SUM(t=1 to infinity) tan^-1(6^r/{2^(2r+1)+3^(2r+1)}) #JEEMAIN2021
Let f(x)=lim n→∞ sum r=0 to n(tan(x/2^(r+1))+tan³(x/2^(r+1))/1-tan²(x/2^(r+1))) Then limx→0(e^x-e^f Let f(x)=lim n→∞ sum r=0 to n(tan(x/2^(r+1))+tan³(x/2^(r+1))/1-tan²(x/2^(r+1))) Then limx→0(e^x-e^f
Let f(x) = lim n tends to infinity sigma r = 0 to r =n ((tan(x/2^(r+1) )+〖tan〗^3 (x/2^(r+1) ))/(1- Let f(x) = lim n tends to infinity sigma r = 0 to r =n ((tan(x/2^(r+1) )+〖tan〗^3 (x/2^(r+1) ))/(1-