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The value of sum sum(n=1)^(oo)cot^(-1)((...

The value of sum `sum_(n=1)^(oo)cot^(-1)(((n^(2)+2n)(n^(2)+2n+1)+1)/(2n+2))` is equal to If `f(x+y)=f(x)+f(y)-x y-1`, `AA x , `y in R ` and `f(1)=1,` then the number of solution of `f(n)=n` , `n in N ,` is

A

`cos^(-1)((1)/(sqrt(5)))`

B

`sec^(-1)((sqrt(5))/(3))`

C

`sin^(-1)((1)/(sqrt(5)))`

D

`cot^(-1)(1)`

Text Solution

Verified by Experts

The correct Answer is:
C

`T_(n)=tan^(-1)((2n+2)/(1+(n^(2)+2n)(n^(2)+2n+1)))`
`=tan^(-1)((2n+2)/(1+n(n+2)(n+1)(n+1)))`
`=(tan^(-1)(n+1)(n+2)-tan^(-1)n(n+1))`
`therefore S_(n)=sum_(n=1)^(n)T_(n)=(tan^(-1)(n+1)(n+2)-tan^(-1)2)`
So, `lim_(n rarr oo)S_(n)=((pi)/(2)-tan^(-1)2)=cot^(-1)2=sin^(-1)((1)/(sqrt(5)))`
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