# Can you help me calculite limit tan^-1(1/3)+tan^-1(1/7)+,,,,,+tan^-1(1/x2+x+1)? x close to infinty?

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You need to use the following trigonometric identity, such that:

`tan^(-1) (a - b)/(1 + ab) = tan^(-1) a- tan^(-1) b`

Considering `a = 1/k`  and `b = 1/(k+1) ` yields:

`tan^(-1) (1/k - 1/(k+1))/(1 + 1/(k(k+1))) = tan^(-1) 1/k - tan^(-1)(1/(k+1))`

`tan^(-1) ((k+1-k)/(k(k+1)))/((k^2 + k + 1)/(k(k+1))) = tan^(-1) 1/k - tan^(-1)(1/(k+1))`

Reducing duplicate factors yields:

`tan^(-1) 1/(k^2 + k + 1) = tan^(-1) 1/k - tan^(-1)(1/(k+1))`

Resoning by analogy yields:

`tan^(-1)(1/3) = tan^(-1) 1/(1^2 + 1 + 1) = tan^(-1) 1 - tan^(-1)1/2`

`tan^(-1) (1/7) = tan^(-1) 1/(2^2 + 2 + 1) = tan^(-1) 1/2 - tan^(-1) 1/3`

Hence, performing the addition of terms yields:

`tan^(-1)(1/3) + tan^(-1)(1/3) + .... + tan^(-1)(1/(x^2 + x +1)) = tan^(-1) 1 - tan^(-1) 1/2 + tan^(-1) 1/2 - tan^(-1) 1/3 + .... + tan^(-1) 1/x - tan^(-1) 1/(x+1)`

Reducing duplicate terms yields:

`tan^(-1)(1/3) + tan^(-1)(1/3) + .... + tan^(-1)(1/(x^2 + x +1)) = tan^(-1) 1 - tan^(-1) 1/(x+1)`

You may evaluate the limit such that:

`lim_(x->oo) (tan^(-1)(1/3) + tan^(-1)(1/3) + .... + tan^(-1)(1/(x^2 + x +1))) = lim_(x->oo) (tan^(-1) 1 - tan^(-1) 1/(x+1))`

`lim_(x->oo) (tan^(-1) 1 - tan^(-1) 1/(x+1)) = tan^(-1) 1 - lim_(x->oo) tan^(-1) 1/(x+1)`

Since `lim_(x->oo) tan^(-1) 1/(x+1) -> tan^(-1) 0`  yields:

`lim_(x->oo) (tan^(-1) 1 - tan^(-1) 1/(x+1)) = pi/4 - tan^(-1) 0`

`lim_(x->oo) (tan^(-1) 1 - tan^(-1) 1/(x+1)) = pi/4 - 0`

`lim_(x->oo) (tan^(-1) 1 - tan^(-1) 1/(x+1)) = pi/4`

Hence, evaluating the given limit, using the trigonometric identity `tan^(-1) (a - b)/(1 + ab) = tan^(-1) a - tan^(-1) b` , yields `lim_(x->oo) (tan^(-1)(1/3) + tan^(-1)(1/3) + .... + tan^(-1)(1/(x^2 + x +1))) = pi/4` .