Let \(\varpi=\dfrac{\Gamma^2\left(\frac14\right)}{2\sqrt{2\pi}}=2.62205755\ldots\) be the lemniscate constant. Then,
\[\Large\displaystyle\sum_{n=1}^\infty\dfrac{1}{\sinh^4(\pi n)}=\dfrac{\varpi^4}{30\pi^4}+\dfrac{1}{3\pi}-\dfrac{11}{90}\]

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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.

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An infinite sum consisting of Fibonacci numbers:

\[\displaystyle\sum_{n\geq0}\binom{2n}{n}\dfrac{F_n}{8^n}=\sqrt{\dfrac{2}{5}}\]

#FibonacciNumber #Fibonacci #FibonacciSequence #FibonacciNumbers #FibonacciSeries #InfiniteSum #InfiniteSeries #Sum #Maths #Math

Interestingly, \(\pi^\text{th}\) root of \(4\) can be expressed as the following limit.
\[\boxed{\displaystyle\sqrt[\pi]4=\lim_{n\to\infty}\prod_{k=0}^n\dfrac{\pi}{2\tan^{-1}(n+k)}}\]

Proof:
Let \(\Phi_n=\displaystyle\prod_{k=0}^n\dfrac{\pi}{2\tan^{-1}(n+k)}\). And taking natural logarithms on both sides, we get
\[\begin{align*}\ln\Phi_n&=\displaystyle\sum_{k=0}^n\left(\ln\left(\frac{\pi}{2}\right)-\ln\tan^{-1}(n+k)\right)\\&=\sum_{k=0}^n\left(\ln\left(\frac{\pi}{2}\right)-\ln\left(\dfrac{\pi}{2}-\tan^{-1}\left(\dfrac{1}{n+k}\right)\right)\right)\\&=-\sum_{k=0}^n\ln\left(1-\dfrac{2}{\pi}\tan^{-1}\left(\dfrac{1}{n+k}\right)\right)\\&=-\sum_{k=0}^n\left(-\dfrac{2}{\pi(n+k)}+O\left(\dfrac{1}{n^2}\right)\right)\\&=\dfrac{2}{\pi}\sum_{k=0}^n\dfrac{1}{n+k}+O\left(\dfrac{1}{n}\right)\end{align*}\]
Taking limit \(n\to\infty\), we get
\[\begin{align*}\displaystyle\lim_{n\to\infty}\ln\Phi_n&=\dfrac{2}{\pi}\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=0}^n\dfrac{1}{1+\frac{k}{n}}+\lim_{n\to\infty}O\left(\dfrac{1}{n}\right)\\&=\dfrac{2}{\pi}\displaystyle\int_0^1\dfrac{\mathrm{d}x}{1+x}\\&=\dfrac{2}{\pi}\ln2=\ln\sqrt[\pi]4\end{align*}\]

Hence, the required limit:
\[\boxed{\displaystyle\lim_{n\to\infty}\prod_{k=0}^n\dfrac{\pi}{2\tan^{-1}(n+k)}=\displaystyle\lim_{n\to\infty}\Phi_n=\sqrt[\pi]4}\ \textbf{Q.E.D.}\ \blacksquare\]

*Note:
\(\tan^{-1}(x)=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}+O(x^6)\)
\(\ln(1-x)=-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-\dfrac{x^4}{4}-\dfrac{x^5}{5}+O(x^6)\)
\(\ln\left(1-\dfrac{2}{\pi}\tan^{-1}(x)\right)=-\dfrac{2x}{\pi}+O(x^2)\)

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How could adding positive integers equal a negative fraction?🤔 🔗 https://www.cantorsparadise.com/the-ramanujan-summation-1-2-3-1-12-a8cc23dea793

Ramanujan's summation: \(1+2+3+\cdots=-\dfrac{1}{12}\)

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