Question

Find the sum

cot-1 2 + cot-1 8 + cot-1 18 +………… to infinity.

Answer 1

Answer:

Let tn denote the nth term of the series

Then tn = cot^-1 2n^2

or, tn = cot^-1 ((4n2-1+1)/2)

= cot^-1 [((2n-1) (2n+1) + 1) / ((2n+1) - (2n-1))]

= cot^-1 (2n – 1) – cot^-1 (2n + 1) …… (1)

putting n = 1, 2, 3, ………… etc. in (1), we get

t1 = cot^-1 1 – cot^-1 3 = cot^-1 2

t2 = cot^-1 3 – cot^-1 5 = cot^-1 8

t3 = cot^-1 5 – cot^-1 7 = cot^-1 18

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tn = cot^-1 (2n – 1) – cot^-1 (2n + 1) = cot^-1 2n2

adding Sn = cot^-1 1 – cot^-1 (2n + 1)

lim n→∞Sn = lim n→∞ (cot^-11 - cot^-1 (2n+1) )

= π /4 – lim n n→∞ cot^-1 (2n+1)

= π /4 – 0 = π /4.

Hence the required sum = π /4.

Hope it helps you....

These sums are in exersice 9.3

I know about these sums