I-heard-this-sum-can-be-close-using-the-Digamma-function-Please-help-me-use-it-i-don-t-know-it-sum-of-nth-term-1-1-2-3-1-4-5-6-1-7-8-9-

Question Number 45316 by Tawa1 last updated on 11/Oct/18

I heard this sum can be close using the Digamma function .

Please help me use it . i {don}^{'} t know it .

sum of nth term : \frac{1}{1.2.3} + \frac{1}{4.5.6} + \frac{1}{7.8.9} + \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

T_{n} = \frac{1}{{1 + (n - 1) 3} {2 + (n - 1) 3} {3 + (n - 1) 3}}

T_{n} = \frac{1}{(3 n - 2) ((3 n - 1) (3 n)}

h e r e i a m u s i n g s o m e f o r m u l a m e n t i o n e d

H a l l a n d k n i g h t A l g e b r a \dots

S_{n} = C - \frac{1}{(3 n - 1) (3 n) \times 2 \times 1}

n = 1

\frac{1}{1 \times 2 \times 3} = C - \frac{1}{2 \times 3 \times 2}

C = \frac{1}{6} + \frac{1}{12} = \frac{2 + 1}{12} = \frac{1}{4}

S_{n} = \frac{1}{4} - \frac{1}{2 (3 n - 1) (3 n)}

r e c h e c k \dots

S_{1} = \frac{1}{4} - \frac{1}{2 \times 2 \times 3} = \frac{3 - 1}{12} = \frac{1}{6} = T_{1} = \frac{1}{1 \times 2 \times 3} = \frac{1}{6}

c o n f o r m

S_{2} = \frac{1}{4} - \frac{1}{2 \times 5 \times 6} = \frac{15 - 1}{60} = \frac{14}{60} = \frac{7}{30}

T_{2} = S_{2} - S_{1} = \frac{7}{30} - \frac{1}{6} = \frac{7 - 5}{30} = \frac{1}{15}

b u t T_{2} = \frac{1}{120}

S o m y a n s w e r i s n o t c o r r e c t \dots

p l s m e n h i o n t h e s o u r c e o f t h i s p r o b l e m

S_{n} = \frac{1}{1 \times 2 \times 3} + \frac{1}{4 \times 5 \times 6} + \frac{1}{7 \times 8 \times 9} + \dots

=

Commented by Tawa1 last updated on 12/Oct/18

The question was asked online .

But sir, when can i apply the formular you used to find the sum .

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

o k . . i a m p o s t i n g \dots

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by Tawa1 last updated on 12/Oct/18

Wow, God bless you sir . I really appreciate your effort .

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