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Question Number 107101 by Dwaipayan Shikari last updated on 08/Aug/20
Fun time    1+2x+3x^2 +4x^3 +....=(1/((1−x)^2 ))  1+4+12+32+...=(1/((1−2)^2 ))  4+12+32+....=0  (No 1 fun)    5+11+17+23+...=0     Σ_(n=1) ^∞ 6n−1=6Σ_(n=1) ^∞ n−Σ^∞ 1=6.(−(1/(12)))−(−(1/2))=0  Σ^∞ n=−(1/(12))    (Ramanujan sum)  Σ^∞ 1=1+1+1+1+1+...=−(1/2)  Σ^∞ n^2 .Σ^∞ (1/n^2 )≥(Σ^∞ 1)^2    (Cauchy schwarz ineqality)  Σ^∞ n^2 .(π^2 /6)≥(1/4)  Σ^∞ n^2 ≥(3/(2π^2 ))
$$\mathrm{Fun}\:\mathrm{time} \\ $$$$ \\ $$$$\mathrm{1}+\mathrm{2x}+\mathrm{3x}^{\mathrm{2}} +\mathrm{4x}^{\mathrm{3}} +….=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$\mathrm{1}+\mathrm{4}+\mathrm{12}+\mathrm{32}+…=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\mathrm{4}+\mathrm{12}+\mathrm{32}+….=\mathrm{0}\:\:\left(\mathrm{No}\:\mathrm{1}\:\mathrm{fun}\right) \\ $$$$ \\ $$$$\mathrm{5}+\mathrm{11}+\mathrm{17}+\mathrm{23}+…=\mathrm{0}\:\:\: \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{6n}−\mathrm{1}=\mathrm{6}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{n}−\overset{\infty} {\sum}\mathrm{1}=\mathrm{6}.\left(−\frac{\mathrm{1}}{\mathrm{12}}\right)−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\overset{\infty} {\sum}\mathrm{n}=−\frac{\mathrm{1}}{\mathrm{12}}\:\:\:\:\left(\mathrm{Ramanujan}\:\mathrm{sum}\right) \\ $$$$\overset{\infty} {\sum}\mathrm{1}=\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+…=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\overset{\infty} {\sum}\mathrm{n}^{\mathrm{2}} .\overset{\infty} {\sum}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\geqslant\left(\overset{\infty} {\sum}\mathrm{1}\right)^{\mathrm{2}} \:\:\:\left(\mathrm{Cauchy}\:\mathrm{schwarz}\:\mathrm{ineqality}\right) \\ $$$$\overset{\infty} {\sum}\mathrm{n}^{\mathrm{2}} .\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\geqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\overset{\infty} {\sum}\mathrm{n}^{\mathrm{2}} \geqslant\frac{\mathrm{3}}{\mathrm{2}\pi^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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