7 days to 2016 #125 Follower Problem

Calculus Level 4

∑ n = 0 ∞ 1 n + 1 ∑ k = 0 n ( n k ) ( − 1 ) k k + 1 = π a b \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n+1 } } \sum _{ k=0 }^{ n }{ \left( \begin{matrix} n \\ k \end{matrix} \right) } \frac { { \left( -1 \right) }^{ k } }{ k+1 } = \frac {\pi^a}b

The equation above holds true for integers a a and b . b. Submit your answer as a + b a + b .


The answer is 8.

Department 8 by Department 8 , 27, Israel

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2 solutions

Naren Bhandari
May 16, 2021

Since ∑ k = 0 n ( n k ) ( − 1 ) k k + 1 = ∫ 0 1 ( ∑ k = 0 n ( n k ) ( − x ) k ) d x = ∫ 0 1 ( 1 − x ) n d x = ∫ 0 1 x n d x = 1 n + 1 \sum_{k=0}^{n}{n\choose k}\frac{(-1)^k}{k+1} =\int_0^1\left(\sum_{k=0}^n{n\choose k} (-x)^k\right)dx=\int_0^1(1-x)^ndx=\int_0^1 x^ndx=\frac{1}{n+1} which give the final closed form ∑ n ≥ 0 1 ( n + 1 ) 2 = ∑ n ≥ 1 1 n 2 = π 2 6 \sum_{n\geq 0}\frac{1}{(n+1)^2}=\sum_{n\geq 1}\frac{1}{n^2}=\frac{\pi^2}{6}

The finite sum can be observed as follows by splitting the sum in even and odd parity f k ≥ 0 ( n ) = ∑ k ≤ n ( n k ) ( − 1 ) k k + 1 = ∑ k ≤ n ( ( n 2 k ) 1 2 k + 1 − ( n 2 k + 1 ) 1 2 k + 2 ) = ∑ k ≤ n ( ( n + 1 2 k + 1 ) − ( n + 1 2 k + 2 ) ) 1 n + 1 = 1 n + 1 f_{k\geq 0}(n)=\sum_{k\leq n}{n\choose k}\frac{(-1)^k}{k+1} =\sum_{k\leq n}\left({n\choose 2k}\frac{1}{2k+1}-{n\choose 2k+1}\frac{1}{2k+2}\right)=\sum_{k\leq n}\left({n+1\choose 2k+1}-{n+1\choose 2k+2}\right)\frac{1}{n+1}=\frac{1}{n+1} Alternative way to go is explained by Carsten Meyer in comment section.

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@Naren Bhandari I'm not sure about the argument with the telescopic sum at the end - shouldn't it be k ≤ n 2 k\leq\boxed{\frac{n}{2}} from the second sum onwards? Also the binomial coefficients don't telescope because one coefficient is odd and the other is even...

It might be easier not to split the sum into even and odd parity and just use the "Binomial Theorem": ∑ k = 0 n ( n k ) ( − 1 ) k k + 1 = 1 n + 1 ∑ k = 0 n ( n + 1 k + 1 ) ( − 1 ) k = 1 n + 1 ( 1 − ∑ k = 0 n + 1 ( n + 1 k ) ( − 1 ) k ) \sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{k+1}=\frac{1}{n+1}\sum_{k=0}^n\binom{n+1}{k+1}(-1)^k=\frac{1}{n+1}\left(1 - \cancel{\sum_{k=0}^{n+1}\binom{n+1}{k}(-1)^k}\right)

Carsten Meyer - 3 weeks, 5 days ago

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Thank you for spotting it out. I didnt check last steps of mine throughly since so I made comment of telescoping( which is a mistake I did). I have updated solution now.

Naren Bhandari - 2 weeks, 2 days ago
Felix Belair
May 18, 2021

I just saw the form π a b \frac{\pi^{a}}{b} which reminded me of the Basel problem so I guessed π 2 6 \frac{\pi^{2}}{6}

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