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Question Number 107101 by Dwaipayan Shikari last updated on 08/Aug/20

$$\mathrm{Fun}\mathrm{time}$$$$$$$$1+2\mathrm{x}+{3\mathrm{x}}^2+{4\mathrm{x}}^3+....=\frac{1}{{(1-\mathrm{x})}^2}$$$$1+4+12+32+...=\frac{1}{{(1-2)}^2}$$$$4+12+32+....=0\left(\mathrm{No}1\mathrm{fun}\right)$$$$$$$$5+11+17+23+...=0$$$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}6\mathrm{n}-1=6\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\mathrm{n}-\sum ^{\mathrm{\infty }}1=6.(-\frac{1}{12})-(-\frac{1}{2})=0$$$$\sum ^{\mathrm{\infty }}\mathrm{n}=-\frac{1}{12}\left(\mathrm{Ramanujan}\mathrm{sum}\right)$$$$\sum ^{\mathrm{\infty }}1=1+1+1+1+1+...=-\frac{1}{2}$$$$\sum ^{\mathrm{\infty }}{\mathrm{n}}^2.\sum ^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2}⩾{\left(\sum ^{\mathrm{\infty }}1\right)}^2\left(\mathrm{Cauchy}\mathrm{schwarz}\mathrm{ineqality}\right)$$$$\sum ^{\mathrm{\infty }}{\mathrm{n}}^2.\frac{{\pi }^2}{6}⩾\frac{1}{4}$$$$\sum ^{\mathrm{\infty }}{\mathrm{n}}^2⩾\frac{3}{2{\pi }^2}$$$$$$$$$$$$$$

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