#binomialcoefficient — Public Fediverse posts
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Try to prove the following two results that relate the harmonic numbers to the golden ratio. Have an excellent weekend.
\[\displaystyle\sum_{n=1}^\infty\binom{2n}n\dfrac{H_n}{5^n}=2\sqrt5\ln\varphi\]
\[\displaystyle\sum_{n=1}^\infty\binom{2n}n\dfrac{H_n}{5^nn}=\frac{2\pi^2}{15}-2\ln^2\varphi\]
where \(\varphi=\frac{1+\sqrt5}2\) is the golden ratio; and \(H_n=\left(1+\frac12+\frac13+\ldots+\frac1n\right)\) is the \(n\)-th harmonic number.
#GoldenRatio #HarmonicNumbers #HarmonicNumber #Logarithm #Pi #Summation #Math #Sum #InfiniteSum #Binomial #BinomialCoefficient #Maths #WeekendChallenge
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Try to prove the following two results that relate the harmonic numbers to the golden ratio. Have an excellent weekend.
\[\displaystyle\sum_{n=1}^\infty\binom{2n}n\dfrac{H_n}{5^n}=2\sqrt5\ln\varphi\]
\[\displaystyle\sum_{n=1}^\infty\binom{2n}n\dfrac{H_n}{5^nn}=\frac{2\pi^2}{15}-2\ln^2\varphi\]
where \(\varphi=\frac{1+\sqrt5}2\) is the golden ratio; and \(H_n=\left(1+\frac12+\frac13+\ldots+\frac1n\right)\) is the \(n\)-th harmonic number.
#GoldenRatio #HarmonicNumbers #HarmonicNumber #Logarithm #Pi #Summation #Math #Sum #InfiniteSum #Binomial #BinomialCoefficient #Maths #WeekendChallenge
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Try to prove the following two results that relate the harmonic numbers to the golden ratio. Have an excellent weekend.
\[\displaystyle\sum_{n=1}^\infty\binom{2n}n\dfrac{H_n}{5^n}=2\sqrt5\ln\varphi\]
\[\displaystyle\sum_{n=1}^\infty\binom{2n}n\dfrac{H_n}{5^nn}=\frac{2\pi^2}{15}-2\ln^2\varphi\]
where \(\varphi=\frac{1+\sqrt5}2\) is the golden ratio; and \(H_n=\left(1+\frac12+\frac13+\ldots+\frac1n\right)\) is the \(n\)-th harmonic number.
#GoldenRatio #HarmonicNumbers #HarmonicNumber #Logarithm #Pi #Summation #Math #Sum #InfiniteSum #Binomial #BinomialCoefficient #Maths #WeekendChallenge
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Try to prove the following two results that relate the harmonic numbers to the golden ratio. Have an excellent weekend.
\[\displaystyle\sum_{n=1}^\infty\binom{2n}n\dfrac{H_n}{5^n}=2\sqrt5\ln\varphi\]
\[\displaystyle\sum_{n=1}^\infty\binom{2n}n\dfrac{H_n}{5^nn}=\frac{2\pi^2}{15}-2\ln^2\varphi\]
where \(\varphi=\frac{1+\sqrt5}2\) is the golden ratio; and \(H_n=\left(1+\frac12+\frac13+\ldots+\frac1n\right)\) is the \(n\)-th harmonic number.
#GoldenRatio #HarmonicNumbers #HarmonicNumber #Logarithm #Pi #Summation #Math #Sum #InfiniteSum #Binomial #BinomialCoefficient #Maths #WeekendChallenge
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Try to prove the following two results that relate the harmonic numbers to the golden ratio. Have an excellent weekend.
\[\displaystyle\sum_{n=1}^\infty\binom{2n}n\dfrac{H_n}{5^n}=2\sqrt5\ln\varphi\]
\[\displaystyle\sum_{n=1}^\infty\binom{2n}n\dfrac{H_n}{5^nn}=\frac{2\pi^2}{15}-2\ln^2\varphi\]
where \(\varphi=\frac{1+\sqrt5}2\) is the golden ratio; and \(H_n=\left(1+\frac12+\frac13+\ldots+\frac1n\right)\) is the \(n\)-th harmonic number.
#GoldenRatio #HarmonicNumbers #HarmonicNumber #Logarithm #Pi #Summation #Math #Sum #InfiniteSum #Binomial #BinomialCoefficient #Maths #WeekendChallenge
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#DifferentialPropositionalCalculus • 4.5
• https://inquiryintoinquiry.com/2020/02/25/differential-propositional-calculus-4/Each of the families — #LinearPropositions, #PositivePropositions, #SingularPrpositions — is naturally parameterized by the coordinate \(n\)-tuples in \(\mathbb{B}^n\) and falls into \(n+1\) ranks, with a #BinomialCoefficient \(\tbinom{n}{k}\) giving the number of propositions having rank or weight \(k\) in their class.
Related Subjects —
#Logic #LogicalGraphs #DifferentialLogic
#PropositionalCalculus #BooleanFunctions