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Question Number 26020 by abdo imad last updated on 17/Dec/17

answer to 25962...we S=Σ_(n=1) ^∝ 1/_(n^2 (n+1)) and S_n = Σ_(k=1) ^(k=n) 1/_(k^2 (k+1)) we have S= lim_(n−>∝) S_n we decompose the the rational fraction F(X)= 1/_X 2_((X^2 +1)) ....F(X)= a/X +b/X^2 + c/X+1 we find a=−1..b=1..c=1 and F(X)= −1/X + 1/X+1+1/X^2 Σ_(k=1) ^(k=n) 1/_(k^2 (k+1)) = −Σ_(k=1) ^(k=n) 1/k + Σ_(k=1) ^(k=n) 1/_(k+1) + Σ_(k=1) ^(k=n) 1/_k^2 but Σ_(k=1) ^(k=n) 1/k = H_n Σ_(k=1) ^(k=n) 1/_(k+1) = Σ_(k=2) ^(k=n+1) 1/k = H_(n+1) −1 −> S_n = H_(n+1) −H_n −1 + Σ_(k=1) ^(k=n) 1/_k^2 but H_n =ln(n)+s+o_1 (1/n)...H_(n+1) = ln(n+1) +s +o_2 (1/n) −> H_(n+1) − H_n =ln((n+1)n^(−1) ) + o_3 (1/n) but lim _(n−>∝) o_3 (1/n)=0 so lim H_(n+1) − H_n =0 lim_ Σ_(k=1) ^(k=n) 1/_k^2 = Σ_(k=1) ^∝ 1/_k^2 =π^2 /6 finally... Σ_(n=1) ^(n=∝) 1/_(n^2 (n+1)) = π^2 /6 −1.

$${answer}\:{to}\:\mathrm{25962}...{we}\:{S}=\sum_{{n}=\mathrm{1}} ^{\propto} \:\mathrm{1}/_{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)} \:{and}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\mathrm{1}/_{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)} \\ $$ $${we}\:{have}\:{S}=\:{lim}_{{n}−>\propto} \:{S}_{{n}} \:{we}\:{decompose}\:{the}\:{the}\:{rational}\:{fraction} \\ $$ $${F}\left({X}\right)=\:\:\:\mathrm{1}/_{{X}} \mathrm{2}_{\left({X}^{\mathrm{2}} +\mathrm{1}\right)} ....{F}\left({X}\right)=\:\:{a}/{X}\:\:+{b}/{X}^{\mathrm{2}} \:\:+\:{c}/{X}+\mathrm{1} \\ $$ $${we}\:{find}\:{a}=−\mathrm{1}..{b}=\mathrm{1}..{c}=\mathrm{1}\:\:{and}\:\:{F}\left({X}\right)=\:\:−\mathrm{1}/{X}\:+\:\mathrm{1}/{X}+\mathrm{1}+\mathrm{1}/{X}^{\mathrm{2}} \\ $$ $$\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\mathrm{1}/_{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)} =\:\:\:−\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\mathrm{1}/{k}\:\:+\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\mathrm{1}/_{{k}+\mathrm{1}} \:+\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\mathrm{1}/_{{k}^{\mathrm{2}} } \\ $$ $${but}\:\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\mathrm{1}/{k}\:\:=\:{H}_{{n}} \\ $$ $$\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\mathrm{1}/_{{k}+\mathrm{1}} =\:\:\sum_{{k}=\mathrm{2}} ^{{k}={n}+\mathrm{1}} \:\mathrm{1}/{k}\:=\:{H}_{{n}+\mathrm{1}} −\mathrm{1} \\ $$ $$−>\:{S}_{{n}} =\:{H}_{{n}+\mathrm{1}} −{H}_{{n}} −\mathrm{1}\:+\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\mathrm{1}/_{{k}^{\mathrm{2}} } \\ $$ $${but}\:\:{H}_{{n}} \:={ln}\left({n}\right)+{s}+{o}_{\mathrm{1}} \left(\mathrm{1}/{n}\right)...{H}_{{n}+\mathrm{1}} =\:{ln}\left({n}+\mathrm{1}\right)\:+{s}\:+{o}_{\mathrm{2}} \left(\mathrm{1}/{n}\right) \\ $$ $$−>\:{H}_{{n}+\mathrm{1}} \:−\:{H}_{{n}} ={ln}\left(\left({n}+\mathrm{1}\right){n}^{−\mathrm{1}} \right)\:\:+\:{o}_{\mathrm{3}} \left(\mathrm{1}/{n}\right) \\ $$ $${but}\:{lim}\:_{{n}−>\propto} {o}_{\mathrm{3}} \left(\mathrm{1}/{n}\right)=\mathrm{0}\:\:{so}\:\:\:{lim}\:\:{H}_{{n}+\mathrm{1}} −\:{H}_{{n}} =\mathrm{0} \\ $$ $${lim}_{} \:\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\mathrm{1}/_{{k}^{\mathrm{2}} } \:\:=\:\sum_{{k}=\mathrm{1}} ^{\propto} \:\mathrm{1}/_{{k}^{\mathrm{2}} } \:\:=\pi^{\mathrm{2}} /\mathrm{6}\:\:{finally}... \\ $$ $$\sum_{{n}=\mathrm{1}} ^{{n}=\propto} \:\:\mathrm{1}/_{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)} \:\:=\:\pi^{\mathrm{2}} /\mathrm{6}\:−\mathrm{1}. \\ $$ $$ \\ $$

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