besnuffelfo

besnuffelfo

Answered

2022-09-21

sum n = 2 ( 1 ) n n 2 + n 2
I am not able to do the telescoping process in the above series. I converted it into the following partial fraction:
n = 2 ( 1 ) n ( n + 2 ) ( n 1 )
But nothing seems to cancel (as usually happens) using the telescoping method. How can I solve the above series? Is there any other method to do the above problem?

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Answer & Explanation

rae2721

rae2721

Expert

2022-09-22Added 8 answers

You need to fully expand the partial fraction.
( 1 ) n ( n + 2 ) ( n 1 ) = ( ( n + 2 ) ( n 1 ) ) × ( 1 ) n 3 × ( n + 2 ) ( n 1 ) = ( n + 2 ) × ( 1 ) n 3 × ( n + 2 ) ( n 1 ) ( n 1 ) × ( 1 ) n 3 × ( n + 2 ) ( n 1 ) = ( 1 ) n 3 × ( n 1 ) ( 1 ) n 3 × ( n + 2 )
Now this can be handled using Alternating Harmonic Series.
The final result, as Mathematica calculated, is 5 + 12 log 2 18

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Conrad Beltran

Conrad Beltran

Expert

2022-09-23Added 1 answers

Writing the nth term of your sum in a form that telescopes can be avoid altogether by converting the problem to a double integral as follows.
Noting that
1 n 1 = 0 1 x n 2 d x and 1 n + 2 = 0 1 y n + 1 d y ,
the sum can be rewritten as
n = 2 ( 1 ) n n 2 + n 2 = n = 2 ( 1 ) n ( n 1 ) ( n + 2 ) (1) = 0 1 0 1 n = 2 ( 1 ) n x n 2 y n + 1 d x d y = 0 1 0 1 y x 2 n = 2 ( x y ) n d x d y (2) = 0 1 0 1 y x 2 x 2 y 2 1 + x y d x d y = 0 1 0 1 y 3 1 + x y d x d y = 0 1 y 2 [ ln ( 1 + x y ) ] 0 1 d y = 0 1 y 2 ln ( 1 + y ) d y (3) = 1 3 ln ( 2 ) 1 3 0 1 y 3 1 + y d y (4) = 1 3 ln ( 2 ) 1 3 0 1 ( y 2 y + 1 1 1 + y ) d y = 1 3 ln ( 2 ) 1 3 [ y 3 3 y 2 2 + y ln ( 1 + y ) ] 0 1 = 2 3 ln ( 2 ) 5 18 .
Explanation
1. Interchanging the summation with the double integration.
2. Summing the series which is geometric.
3. Integration by parts.
4. Partial fraction decomposition.

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