Value of limn→∞∑r=1ncot−1r4−r2+12ris

Sn=∑1n tan−1⁡2r1+r4−r2 =∑r=1n tan−1⁡2r1+r2−rr2+r=∑r=1n tan−1⁡r2+r−r2−r1+r2+rr2−r =∑r=1n tan−1⁡r2+r−tan−1⁡r2−rSn=tan−1⁡n2+n−tan−1⁡0limx→∞ Sn=limn→∞ tan−1⁡n2+n=π2

Value of limn→∞∑r=1ncot−1r4−r2+12ris

Sn=∑1n tan−1⁡2r1+r4−r2 =∑r=1n tan−1⁡2r1+r2−rr2+r=∑r=1n tan−1⁡r2+r−r2−r1+r2+rr2−r =∑r=1n tan−1⁡r2+r−tan−1⁡r2−rSn=tan−1⁡n2+n−tan−1⁡0limx→∞ Sn=limn→∞ tan−1⁡n2+n=π2

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Value of $\lim_{n \to \infty} \sum_{r = 1}^{n} \cot^{- 1} (\frac{r^{4} - r^{2} + 1}{2 r}) is$

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$π$

$\frac{π}{4}$

$\frac{3 π}{4}$

$\frac{π}{2}$

answer is B.

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Detailed Solution

$S_{n} = \sum_{1}^{n} \tan^{- 1} \frac{2 r}{1 + r^{4} - r^{2}}$ $\begin{aligned} = \sum_{r = 1}^{n} \tan^{- 1} \frac{2 r}{1 + (r^{2} - r) (r^{2} + r)} \\ = \sum_{r = 1}^{n} \tan^{- 1} (\frac{r^{2} + r - (r^{2} - r)}{1 + (r^{2} + r) (r^{2} - r)}) \end{aligned}$ $\begin{aligned} = \sum_{r = 1}^{n} \tan^{- 1} (r^{2} + r) - \tan^{- 1} (r^{2} - r) \\ S_{n} = \tan^{- 1} (n^{2} + n) - \tan^{- 1} 0 \\ lim_{x \to \infty} S_{n} = lim_{n \to \infty} \tan^{- 1} (n^{2} + n) = \frac{π}{2} \end{aligned}$