Inverse Sum Quintic
n = 1 ∑ ∞ ⎝ ⎛ k = 0 ∑ n k 5 ⎠ ⎞ − 1
If the value of the series above can be expressed as A π 2 + B π tan ( 2 π C ) + D where A , B , C and D are integers, find the value of A + B + C + D .
The answer is 251.
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1 solution
Where did the tan(pi * sqrt3 /2) come from?
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To do ∑ 2 n 2 + 2 n − 1 1 = ∑ 2 ( n + a ) ( n + b ) 1 we can use ∑ ( n + a ) ( n + b ) 1 = b − a ψ ( b + 1 ) − ψ ( a + 1 ) , where ψ is the digamma function. Using the recursion and the reflection properties of the digamma function, we can work it out ... details left to the reader as an exercise ;)
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You did Digamma? I didn't. This is the most important part of the solution that you're missing!!!! Can you please fill in the gaps?
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@Pi Han Goh – Will write it up over the weekend... I'm a busy calculus teacher now ;)
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@Otto Bretscher – Note to self: Remind Mr. Otto Bretscher to post a solution in 36 hours time.
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@Pi Han Goh – We have a , b = 2 1 ± 3 , and the "most important part of the solution" (as you put it) is computing ψ ( b + 1 ) − ψ ( a + 1 ) . Using first the recursion formula and then the reflection formula (since a + b = 1 ) for digamma, we find ψ ( b + 1 ) − ψ ( a + 1 ) = ψ ( b ) − ψ ( a ) + b 1 − a 1 = π cot ( π a ) − 2 3 = − π tan ( 2 3 π ) − 2 3 which is what we want. (The details are just bookkeeping)
I too used digamma. They are the most general way of doing the "products in denominator" problems. BTW, what about n > 5 or so? Can we have a general answer? Well, that would depend if we can factorize the Faullhaber's formula's polynomial. Can we? Or if one can do it without factorizing? Well, as far as I know, there is not. But still let me confirm.
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@Kartik Sharma – Answer = no. Unless you can keep them in digamma forms. Even powers of 4 is not a possibility.
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@Pi Han Goh – Oh, I see. Thanks! Nice problem!
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@Kartik Sharma – Can you post a solution? ahhaha. Possibly a non-English one.
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@Pi Han Goh – I don't think I can and I don't think I should. Because what I have done is same as Sir Otto has done and he has explained it very well. Rather, I am waiting for your solution(without digamma).
@Pi Han Goh – "non-English" solution? ;)
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@Otto Bretscher – Basically solutions that are way too hard for most people to understand. See Kartik's solution and my comment here .
Lovely and over the top!
The inside term is n 2 ( n + 1 ) 2 ( 2 n 2 + 2 n − 1 ) 1 2 = − n 2 1 2 − ( n + 1 ) 2 1 2 + 2 n 2 + 2 n − 1 4 8 , and the sum works out to − 2 π 2 − ( 2 π 2 − 1 2 ) + 8 ( 6 + 3 π tan ( 2 3 π ) ) . Details omitted ;)