Find-the-sum-to-close-form-1-1-2-3-1-4-5-6-1-7-8-9-

Question Number 147877 by Tawa11 last updated on 24/Jul/21

Find the sum to close form .

\frac{1}{1.2.3} + \frac{1}{4.5.6} + \frac{1}{7.8.9} + \dots

Commented by Tinku Tara last updated on 24/Jul/21

Welcome back . Tawa .

Commented by Tawa11 last updated on 24/Jul/21

Thank you sir . (\overset{⌢ . . ⌢}{⌣})

Answered by Olaf_Thorendsen last updated on 24/Jul/21

S = \underset{n = 0}{\sum^{\infty}} \frac{1}{(3 n + 1) (3 n + 2) (3 n + 3)}

S = \frac{1}{8} \underset{n = 0}{\sum^{\infty}} \frac{3}{(3 n + 1) (3 n + 2)} - \frac{5}{(3 n + 2) (3 n + 3)} + \frac{2}{(3 n + 3) (3 n + 1)}

S = \frac{1}{24} \underset{n = 0}{\sum^{\infty}} \frac{1}{(n + \frac{1}{3}) (n + \frac{2}{3})} - \frac{5}{72} \underset{n = 0}{\sum^{\infty}} \frac{1}{(n + \frac{2}{3}) (n + 1)}

+ \frac{1}{36} \underset{n = 0}{\sum^{\infty}} \frac{1}{(n + 1) (n + \frac{1}{3})}

S = \frac{1}{24} . \frac{ψ (\frac{2}{3}) - ψ (\frac{1}{3})}{\frac{2}{3} - \frac{1}{3}} - \frac{5}{72} . \frac{ψ (1) - ψ (\frac{2}{3})}{1 - \frac{2}{3}}

+ \frac{1}{36} . \frac{ψ (1) - ψ (\frac{1}{3})}{1 - \frac{1}{3}}

S = \frac{1}{8} (ψ (\frac{2}{3}) - ψ (\frac{1}{3})) - \frac{5}{24} (ψ (1) - ψ (\frac{2}{3}))

+ \frac{1}{24} (ψ (1) - ψ (\frac{1}{3}))

S = - \frac{1}{6} ψ (\frac{1}{3}) + \frac{1}{3} ψ (\frac{2}{3}) - \frac{1}{6} ψ (1)

∙ ψ (1 - z) = ψ (z) + π \cot (π z)

\Rightarrow ψ (\frac{2}{3}) = ψ (\frac{1}{3}) + π \cot (\frac{π}{3})

ψ (\frac{2}{3}) = ψ (\frac{1}{3}) + \frac{π}{\sqrt{3}}

S = \frac{1}{6} ψ (\frac{1}{3}) + \frac{π}{3 \sqrt{3}} - \frac{1}{6} ψ (1)

∙ ψ (1) = - γ

∙ ψ (\frac{1}{3}) = - \frac{π}{2 \sqrt{3}} - \frac{3}{2} \ln 3 - γ

S = \frac{1}{6} (- \frac{π}{2 \sqrt{3}} - \frac{3}{2} \ln 3 - γ) + \frac{π}{3 \sqrt{3}} + \frac{γ}{6}

S = \frac{π}{4 \sqrt{3}} - \frac{1}{4} \ln 3

Commented by Tawa11 last updated on 24/Jul/21

Wow, God bless you sir .

Commented by Tawa11 last updated on 24/Jul/21

Sir, where can I learn summation like this . Please sir prescribe book . I want

to study on it too . Thanks sir .

Commented by Tawa11 last updated on 24/Jul/21

Sir .

Can we find sum to n terms here : \frac{1}{1.2.3} + \frac{1}{4.5.6} + \frac{1}{7.8.9} + \dots

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