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Question Number 147873 by 0731619 last updated on 24/Jul/21

Answered by Olaf_Thorendsen last updated on 24/Jul/21

$$\mathrm{S}=\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{3n+2}-\frac{1}{3n+4})$$$$\mathrm{S}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2}{(3n+2)(3n+4)}$$$$\mathrm{S}=\frac{2}{9}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(n+\frac{2}{3})(n+\frac{4}{3})}$$$$\mathrm{S}=\frac{2}{9}.\frac{\psi \left(\frac{4}{3}\right)-\psi \left(\frac{2}{3}\right)}{\frac{4}{3}-\frac{2}{3}}\left(1\right)$$$$$$$$\bullet \psi (z+1)=\psi \left(z\right)+\frac{1}{z}$$$$\Rightarrow \psi \left(\frac{4}{3}\right)=\psi \left(\frac{1}{3}\right)+3$$$$$$$$\bullet \psi (1-z)=\psi \left(z\right)+\pi \cot \left(\pi z\right)$$$$\Rightarrow \psi \left(\frac{2}{3}\right)=\psi \left(\frac{1}{3}\right)+\pi \cot \left(\frac{\pi }{3}\right)$$$$\psi \left(\frac{2}{3}\right)=\psi \left(\frac{1}{3}\right)+\frac{\pi }{\sqrt{3}}$$$$$$$$\left(1\right):\mathrm{S}=\frac{2}{9}.\frac{(\psi \left(\frac{1}{3}\right)+3)-(\psi \left(\frac{1}{3}\right)+\frac{\pi }{\sqrt{3}})}{\frac{2}{3}}$$$$\mathrm{S}=\frac{2}{9}\times \frac{3}{2}(3-\frac{\pi }{\sqrt{3}})=1-\frac{\pi }{3\sqrt{3}}$$

Commented by Tawa11 last updated on 24/Jul/21

$$\mathrm{Great}$$

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