displaystyle∑n=3∞ tan -1((4/n2-2 n-2)) is equal to

Question

displaystyle∑n=3∞ tan -1((4/n2-2 n-2)) is equal to (A) 3 π (B) 2 π (C) π (D) (π/2)

Tardigrade Admin · Accepted Answer

displaystyle∑n=3∞ tan -1((4/n2-2 n-2))= displaystyle∑n=3∞ tan -1((4/1+(n+1)(n-3)))= displaystyle∑n=3∞( tan -1(n+1)- tan -1(n-3)) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/0fd1abe5e4181af8ce7f91d846396dbd-.png" /> S∞ =2 π - π = π

Q. $n = 3 \sum \infty tan^{- 1} (\frac{4}{n ^{2} - 2 n - 2})$ is equal to

$3 π$

$2 π$

$π$

$\frac{π}{2}$

$n = 3 \sum \infty tan^{- 1} (\frac{4}{n ^{2} - 2 n - 2}) = n = 3 \sum \infty tan^{- 1} (\frac{4}{1 + ( n + 1 ) ( n - 3 )}) = n = 3 \sum \infty (tan^{- 1} (n + 1) - tan^{- 1} (n - 3))$

$S_{\infty} = 2 π - π = π$

Q. n=3∑∞​tan−1(n2−2n−24​) is equal to

3π

2π

π

2π​

n=3∑∞​tan−1(n2−2n−24​)=n=3∑∞​tan−1(1+(n+1)(n−3)4​)=n=3∑∞​(tan−1(n+1)−tan−1(n−3)) S∞​=2π−π=π

Q. $n = 3 \sum \infty tan^{- 1} (\frac{4}{n ^{2} - 2 n - 2})$ is equal to

$3 π$

$2 π$

$π$

$\frac{π}{2}$

$n = 3 \sum \infty tan^{- 1} (\frac{4}{n ^{2} - 2 n - 2}) = n = 3 \sum \infty tan^{- 1} (\frac{4}{1 + ( n + 1 ) ( n - 3 )}) = n = 3 \sum \infty (tan^{- 1} (n + 1) - tan^{- 1} (n - 3))$

$S_{\infty} = 2 π - π = π$