Who's up to the challenge?18

Calculus Level 5

∑ k = 1 ∞ ∑ n = 2 k − 1 + 1 2 k k 2 n ( 2 n − 1 ) \large \displaystyle\sum _{ k=1 }^{ \infty }{ \displaystyle\sum _{ n={ 2 }^{ k-1 }+1 }^{ { 2 }^{ k } }{ \frac { k }{ 2n(2n-1) } } }

If the value of the series above is equal to A − B γ A-B\gamma , where A A and B B are positive integers, find A + B A+B .

Notation : γ \gamma denotes the Euler-Mascheroni constant .

This is a part of Who's up to the challenge?


The answer is 2.

Hamza A by Hamza A , 19, USA

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1 solution

Thomas Welle
Dec 6, 2019

Let's define: H n = ∑ k = 1 n 1 k = 1 + 1 2 + 1 3 + 1 4 + ⋯ + 1 n H_n = \sum^{n}_{k=1} \frac{1}{k} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n} A n = ∑ k = 1 n ( − 1 ) k + 1 k = 1 − 1 2 + 1 3 − 1 4 + ⋯ ± 1 n A_n = \sum^{n}_{k=1} \frac{(-1)^{k+1}}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \pm \frac{1}{n} For an even number 2 n 2n , A 2 n = 1 − 1 2 + 1 3 − 1 4 + ⋯ − 1 2 n = ( 1 + 1 2 + 1 3 + 1 4 + ⋯ + 1 2 n ) − 2 ( 1 2 + 1 6 + 1 8 + ⋯ + 1 2 n ) = H 2 n − H n A_{2n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots - \frac{1}{2n} = \left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{2n} \right) - 2\left( \frac{1}{2} + \frac{1}{6} + \frac{1}{8} + \dots + \frac{1}{2n} \right) = H_{2n} -H_{n} Also note that, lim ⁡ n → ∞ A n = 1 − 1 2 + 1 3 − 1 4 + ⋯ = ln ⁡ 2 \lim_{n \to \infty} A_{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots=\ln{2}

Now let's let S S denote the sum of this series. S = ∑ k = 1 ∞ ∑ n = 2 k − 1 − 1 2 k k 2 n ( 2 n − 1 ) S= \sum_{k=1}^{\infty} \sum_{n=2^{k-1} -1}^{2^k} \frac{k}{2n(2n-1)} The fraction 1 2 n ( 2 n − 1 ) \frac{1}{2n(2n-1)} can be written as 1 2 n − 1 − 1 2 n \frac{1}{2n-1}-\frac{1}{2n} . So, S = ∑ k = 1 ∞ k ∑ n = 2 k − 1 − 1 2 k 1 2 n − 1 − 1 2 n = 1 ( 1 3 − 1 4 ) + 2 ( 1 5 − 1 6 + 1 7 − 1 8 ) + 3 ( 1 9 − 1 10 + 1 11 − 1 12 + 1 13 − 1 14 + 1 15 − 1 16 ) + … S= \sum_{k=1}^{\infty} k \sum_{n=2^{k-1} -1}^{2^k} \frac{1}{2n-1}-\frac{1}{2n} = 1\left(\frac{1}{3} -\frac{1}{4}\right) +2\left(\frac{1}{5} -\frac{1}{6} +\frac{1}{7} -\frac{1}{8}\right) + 3\left(\frac{1}{9} -\frac{1}{10}+\frac{1}{11} -\frac{1}{12}+\frac{1}{13} -\frac{1}{14}+\frac{1}{15} -\frac{1}{16}\right) +\dots Add the equation 0 = ln ⁡ 2 − ( 1 − 1 2 + 1 3 − 1 4 + … ) 0=\ln2 - \left(1-\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \right) . S = ln ⁡ 2 − ( 1 − 1 2 ) + 1 ( 1 5 − 1 6 + 1 7 − 1 8 ) + 2 ( 1 9 − 1 10 + 1 11 − 1 12 + 1 13 − 1 14 + 1 15 − 1 16 ) + … S=\ln2 -\left(1-\frac{1}{2}\right) +1\left(\frac{1}{5} -\frac{1}{6} +\frac{1}{7} -\frac{1}{8}\right) + 2\left(\frac{1}{9} -\frac{1}{10}+\frac{1}{11} -\frac{1}{12}+\frac{1}{13} -\frac{1}{14}+\frac{1}{15} -\frac{1}{16}\right) +\dots Move ( 1 − 1 2 ) \left(1-\frac{1}{2}\right) to the left hand side, and we can see that we have reduced the coefficient on each of the terms by one. S + ( 1 − 1 2 ) = ln ⁡ 2 + 1 ( 1 5 − 1 6 + 1 7 − 1 8 ) + 2 ( 1 9 − 1 10 + 1 11 − 1 12 + 1 13 − 1 14 + 1 15 − 1 16 ) + … S +\left(1-\frac{1}{2}\right)=\ln2 +1\left(\frac{1}{5} -\frac{1}{6} +\frac{1}{7} -\frac{1}{8}\right) + 2\left(\frac{1}{9} -\frac{1}{10}+\frac{1}{11} -\frac{1}{12}+\frac{1}{13} -\frac{1}{14}+\frac{1}{15} -\frac{1}{16}\right) +\dots Let's repeat this. Add 0 = ln ⁡ 2 − ( 1 − 1 2 + 1 3 − 1 4 + … ) 0=\ln2 - \left(1-\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \right) . S + ( 1 − 1 2 ) = 2 ln ⁡ 2 − ( 1 − 1 2 + 1 3 − 1 4 ) + 1 ( 1 9 − 1 10 + 1 11 − 1 12 + 1 13 − 1 14 + 1 15 − 1 16 ) + … S +\left(1-\frac{1}{2}\right)=2 \ln2 -\left(1 -\frac{1}{2} + \frac{1}{3} - \frac{1}{4} \right) + 1\left(\frac{1}{9} -\frac{1}{10}+\frac{1}{11} -\frac{1}{12}+\frac{1}{13} -\frac{1}{14}+\frac{1}{15} -\frac{1}{16}\right) +\dots S + ( 1 − 1 2 ) + ( 1 − 1 2 + 1 3 − 1 4 ) = 2 ln ⁡ 2 + 1 ( 1 9 − 1 10 + 1 11 − 1 12 + 1 13 − 1 14 + 1 15 − 1 16 ) + … S +\left(1-\frac{1}{2}\right) +\left(1 -\frac{1}{2} + \frac{1}{3} - \frac{1}{4} \right) =2 \ln2 + 1\left(\frac{1}{9} -\frac{1}{10}+\frac{1}{11} -\frac{1}{12}+\frac{1}{13} -\frac{1}{14}+\frac{1}{15} -\frac{1}{16}\right) +\dots If we repeat this procedure n times, we'll get: S + A 2 + A 4 + A 8 + ⋯ + A 2 n = n ln ⁡ 2 + 1 ( 1 2 n + 1 + 1 − 1 2 n + 1 + 2 + ⋯ − 1 2 n + 2 ) + 2 ( … ) + … S+A_2+A_4+A_8+\dots+A_{2^n}=n\ln2 +1\left(\frac{1}{2^{n+1}+1} -\frac{1}{2^{n+1}+2} +\dots -\frac{1}{2^{n+2}} \right) +2\left( \dots \right) + \dots One can convince oneself that the terms on the right of n ln ⁡ 2 n \ln2 approach 0 in the limit of large n. What about the left hand side? Remember, A 2 n = H 2 n − H n A_{2n} = H_{2n} -H_{n} . So we get: S + ( H 2 − H 1 ) + ( H 4 − H 2 ) + ( H 8 − H 4 ) + ⋯ + ( H 2 n − H 2 n − 1 ) = n ln ⁡ 2 + 1 ( 1 2 n + 1 + 1 − 1 2 n + 1 + 2 + ⋯ − 1 2 n + 2 ) + … S+(H_2-H_1) + (H_4-H_2) +(H_8 -H_4) + \dots+(H_{2^n} -H_{2^{n-1}})=n\ln2 +1\left(\frac{1}{2^{n+1}+1} -\frac{1}{2^{n+1}+2} +\dots -\frac{1}{2^{n+2}} \right) + \dots S − H 1 + H 2 n = n ln ⁡ 2 + 1 ( 1 2 n + 1 + 1 − 1 2 n + 1 + 2 + ⋯ − 1 2 n + 2 ) + … S-H_1 +H_{2^n} =n\ln2+ 1\left(\frac{1}{2^{n+1}+1} -\frac{1}{2^{n+1}+2} +\dots -\frac{1}{2^{n+2}} \right) + \dots S = 1 + ln ⁡ 2 n − H 2 n + 1 ( 1 2 n + 1 + 1 − 1 2 n + 1 + 2 + ⋯ − 1 2 n + 2 ) + … S=1 +\ln2^n -H_{2^n} + 1\left(\frac{1}{2^{n+1}+1} -\frac{1}{2^{n+1}+2} +\dots -\frac{1}{2^{n+2}} \right) + \dots The Euler-Mascheroni constant is defined by γ = lim ⁡ N → ∞ ( H N − ln ⁡ N ) \gamma = \lim_{N \to \infty} (H_{N}-\ln N ) So, taking the limit as n tends to infinity, we get: S = 1 − γ S=1-\gamma and A + B = 2 A+B =2

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