nice-calculus-find-the-value-of-n-1-6n-2-5n-1-

Question Number 138821 by mnjuly1970 last updated on 18/Apr/21

\dots \dots . n i c e . . . . . . c a l c u l u s \dots \dots

f i n d t h e v a l u e o f :

Θ = \underset{n = - \infty}{\sum^{\infty}} \frac{1}{6 n^{2} + 5 n + 1} = ?

Commented by Kamel last updated on 19/Apr/21

Y o u c a n u s e d i g a m m a f u n c t i o n o r r e s i d u e t h e o r e m .

Answered by Kamel last updated on 19/Apr/21

Θ = \underset{n = - \infty}{\sum^{+ \infty}} \frac{1}{6 n^{2} + 5 n + 1} = \frac{1}{6} \underset{n = - \infty}{\sum^{+ \infty}} \frac{1}{(n + \frac{1}{2}) (n + \frac{1}{3})}

= \underset{N \to + \infty}{l i m} \underset{n = 0}{\sum^{N}} (\frac{1}{n + \frac{1}{3}} - \frac{\overset{}{1}}{n + \frac{1}{2}} + \frac{1}{- n + \frac{1}{3}} - \frac{1}{- n + \frac{1}{2}}) - 1

= \underset{N \to + \infty}{l i m} \underset{n = 0}{\sum^{N}} {3 (\frac{1}{3 n + 1} - \frac{1}{3 n - 1}) - 2 (\frac{1}{2 n + 1} - \frac{1}{2 n - 1})} - 1

= \underset{N \to + \infty}{l i m} \underset{n = 0}{\sum^{N - 1}} 3 (\frac{1}{3 n + 1} - \frac{1}{3 n + 2}) = 3 \underset{n = 0}{\sum^{+ \infty}} \frac{1}{(3 n + 1) (3 n + 2)}

P u t : f (x) = \underset{n = 0}{\sum^{+ \infty}} \frac{x^{3 n + 2}}{(3 n + 1) (3 n + 2)}, 0 ⩽ x < 1 .

f^{″} (x) = \underset{n = 0}{\sum^{+ \infty}} x^{3 n} = \frac{1}{1 - x^{3}} = \frac{1}{(1 - x) (1 + x + x^{2})} = \frac{1}{3} (\frac{1}{1 - x} + \frac{2 x + 1 + 3}{2 (x^{2} + x + 1)})

∴ f^{'} (x) + C = \frac{1}{3} (- L n (1 - x) + \frac{1}{2} L n (x^{2} + x + 1) + \sqrt{3} A r c t a n (\frac{2}{\sqrt{3}} (x + \frac{1}{2}))

f^{'} (0) = 0 \Rightarrow C = \frac{π}{6 \sqrt{3}}

f (1) = - \frac{1}{3} \int_{0}^{1} L n (x) d x + \frac{1}{6} \int_{0}^{1} L n (x^{2} + x + 1) d x + \frac{1}{\sqrt{3}} \int_{0}^{1} A r c t a n (\frac{2}{\sqrt{3}} (x + \frac{1}{2})) d x - \frac{π}{6 \sqrt{3}}

I = \int L n (x^{2} + x + 1) d x = x L n (x^{2} + x + 1) - \int \frac{2 (x^{2} + x + 1) - x - 2}{x^{2} + x + 1} d x

= x L n (x^{2} + x + 1) - 2 x + \int \frac{x + 2}{x^{2} + x + 1} d x

= (x + \frac{1}{2}) L n (x^{2} + x + 1) - 2 x + \sqrt{3} A r c t a n (\frac{2}{\sqrt{3}} (x + \frac{1}{2}))

\int A r c t a n (x) d x = x A r c t a n (x) - \frac{1}{2} L n (1 + x^{2})

S o : f (1) = \frac{1}{3} + \frac{1}{6} (\frac{3}{2} L n (3) - 2 + \frac{π \sqrt{3}}{3} - \frac{π \sqrt{3}}{6}) + \frac{1}{2} (\frac{π}{\sqrt{3}} - \frac{π}{6 \sqrt{3}} - \frac{1}{2} L n (3)) - \frac{π}{6 \sqrt{3}}

= \frac{π \sqrt{3}}{36} + \frac{5 π}{12 \sqrt{3}} - \frac{π}{6 \sqrt{3}} = \frac{π}{3 \sqrt{3}}, Θ = 3 f (1)

∴ Θ = \frac{π}{\sqrt{3}} \Rightarrow \underset{n = - \infty}{\sum^{+ \infty}} \frac{1}{6 n^{2} + 5 n + 1} = \frac{π \sqrt{3}}{3}

B E N A I C H A K a m e l

Commented by mnjuly1970 last updated on 19/Apr/21

t h a n k s a l o t m r K m e l \dots

Answered by Dwaipayan Shikari last updated on 19/Apr/21

\underset{n = - \infty}{\sum^{\infty}} \frac{1}{({a n}^{2} + b n + c)} = \frac{1}{\sqrt{b^{2} - 4 a c}} (ψ (\frac{b + \sqrt{b^{2} - 4 a c}}{2}) - ψ (\frac{b - \sqrt{b^{2} - 4 a c}}{2})) + Λ

Λ = \frac{1}{\sqrt{b^{2} - 4 a c}} (ψ (1 - \frac{b}{2 a} + \frac{\sqrt{b^{2} - 4 a c}}{2 a}) - ψ (1 - \frac{b}{2 a} - \frac{\sqrt{b^{2} - 4 a c}}{2 a}))

\underset{n = - \infty}{\sum^{\infty}} \frac{1}{(6 n^{2} + 5 n + 1)} = ψ (\frac{1}{2}) - ψ (\frac{1}{3}) + ψ (\frac{2}{3}) - ψ (\frac{1}{2})

= π c o t (\frac{π}{3}) = \frac{π}{\sqrt{3}}

Commented by mnjuly1970 last updated on 19/Apr/21

t h a n k y o u m r P a y a n \dots

t h a n k y o u m r P a y a n \dots

Answered by mnjuly1970 last updated on 19/Apr/21