Interesting Infinite Sums

Here are some interesting values of infinite sums that I calculated:

\[\displaystyle \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)} = \frac{\pi}{3\sqrt{3}}\]

n=01(4n+1)(4n+3)=π8\displaystyle \sum_{n=0}^{\infty} \frac{1}{(4n+1)(4n+3)} = \frac{\pi}{8} (Leibniz's formula)

n=01(6n+1)(6n+5)=π83\displaystyle \sum_{n=0}^{\infty} \frac{1}{(6n+1)(6n+5)} = \frac{\pi}{8\sqrt{3}}

n=01(8n+1)(8n+7)=π48(2+1)\displaystyle \sum_{n=0}^{\infty} \frac{1}{(8n+1)(8n+7)} = \frac{\pi}{48}(\sqrt{2}+1)

n=01(8n+3)(8n+5)=π16(21)\displaystyle \sum_{n=0}^{\infty} \frac{1}{(8n+3)(8n+5)} = \frac{\pi}{16}(\sqrt{2}-1)

My method for calculation using the result:

r=1ncosrx=sin(n+12)xsin12x2sin12x\displaystyle \sum_{r=1}^n \cos rx = \frac{\sin(n+\frac{1}{2})x-\sin\frac{1}{2}x}{2\sin\frac{1}{2}x} for sin12x0\sin\frac{1}{2}x\neq0

Integrating both sides we get r=1n1rsinrx=sin(n+12)xsin12x2sin12xdx\displaystyle \sum_{r=1}^n \frac{1}{r}\sin rx = \int \frac{\sin(n+\frac{1}{2})x-\sin\frac{1}{2}x}{2\sin\frac{1}{2}x} dx

Simplifying the integral as making the appropriate substitution u=12xu=\frac{1}{2}x we get the result:

x+r=1n1rsin2rx=sin(2n+1)xsinxdx x + \displaystyle \sum_{r=1}^n \frac{1}{r}\sin 2rx = \int \frac{\sin (2n+1)x}{\sin x} dx

Using the fact that limn0asin(2n+1)xsinxdx=π2\lim_{n\to\infty} \int_{0}^{a} \frac{\sin (2n+1)x}{\sin x} dx = \frac{\pi}{2} for positive aa and aqπa\neq q\pi where qq is a positive integer, we can establish infinite sums such as:

π6+r=11rsinrπ3=π2 \frac{\pi}{6} + \displaystyle \sum_{r=1}^{\infty} \frac{1}{r}\sin \frac{r\pi}{3} = \frac{\pi}{2}

π8+r=11rsinrπ4=π2 \frac{\pi}{8} + \displaystyle \sum_{r=1}^{\infty} \frac{1}{r}\sin \frac{r\pi}{4} = \frac{\pi}{2}

π3+r=11rsin2rπ3=π2 \frac{\pi}{3} + \displaystyle \sum_{r=1}^{\infty} \frac{1}{r}\sin \frac{2r\pi}{3} = \frac{\pi}{2}

π4+r=11rsinrπ2=π2 \frac{\pi}{4} + \displaystyle \sum_{r=1}^{\infty} \frac{1}{r}\sin \frac{r\pi}{2} = \frac{\pi}{2}

Note by Chris Sapiano
1 year, 1 month ago

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The Digamma function you see is simply the logarithmic derivative of the gamma function that is take the natural log of the gamma function and take its derivative respect to the input.You can check the more clear definition at Brilliant website .You can also look at my discussion at the calculus section titled 'On the Iterated Properties of Digamma function'.Also it's one of my favourite function.

Aruna Yumlembam - 1 year ago

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Thanks,well now I can find some new results regarding Digamma function.

Aruna Yumlembam - 1 year ago

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Can you show them? I have no idea what the digamma function is

Chris Sapiano - 1 year ago

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Hey just wanted to know that the above 5 sums you calculated by the Digamma function or not.If not please send me a different solution .Also I proved all 5 sums via digamma function.

Aruna Yumlembam - 1 year ago

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No i wrote how i proved it underneath

Chris Sapiano - 1 year ago

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Niceee!!! I even liked solving your recent problems!!!

Aaghaz Mahajan - 1 year, 1 month ago

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