Find-lim-n-n-pi-2-4-a-n-2-where-a-n-k-1-n-arctan-1-k-2-k-1-

Question Number 153571 by mathdanisur last updated on 08/Sep/21

Find \underset{n \to \infty}{\lim n} \cdot (\frac{π^{2}}{4} - a_{n}^{2}) = ?

where a_{n} = \underset{k = 1}{\sum^{n}} \arctan (\frac{1}{k^{2} - k + 1})

Answered by mnjuly1970 last updated on 08/Sep/21

s o l u t i o n . .

a_{n} = \underset{k = 1}{\sum^{n}} a r c t a n (\frac{k - (k - 1)}{1 + k (k - 1)})

= \underset{k = 1}{\sum^{n}} {a r c t a n (k) - a r c t a n (k - 1)}

= a r c t a n (n)

∴ {l i m}_{n \to \infty} n . (\frac{π^{2}}{4} - {a r c t a n}^{2} (n))

= {l i m}_{n \to \infty} n . (\frac{π}{2} - a r c t a n (n)) (\frac{π}{2} + a r c t a n (n))

= {l i m}_{n \to \infty} n . a r c c o t (n) . {l i m}_{n \to \infty} (\frac{π}{2} + a r c t a n (n))

= π {l i m}_{n \to \infty} n . a r c c o t (n) = π . 1 = π

Commented by mathdanisur last updated on 08/Sep/21

S er, = \infty or π . ?

Commented by mathdanisur last updated on 08/Sep/21

S er, \arctan (n) or arccot (n) . ?

Commented by mnjuly1970 last updated on 08/Sep/21

a r c c o t a n (n)

Commented by mathdanisur last updated on 08/Sep/21

Very nice thankyou ser