Find-lim-n-k-1-n-i-1-k-i-k-i-1-3-1-

Question Number 218365 by hardmath last updated on 08/Apr/25

Find: 𝛀 =lim_(n→∞) Σ_(k=1) ^n [ Σ_(i=1) ^k i (k − i + (1/3))]^(−1) = ?

$$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\left[\:\underset{\boldsymbol{\mathrm{i}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{k}}} {\sum}}\:\mathrm{i}\:\left(\mathrm{k}\:−\:\mathrm{i}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\right)\right]^{−\mathrm{1}} =\:? \\ $$

Answered by vnm last updated on 08/Apr/25

Σ_(i=1) ^k i(k−i+(1/3))=(k+(1/3))Σ_(i=1) ^k i−Σ_(i=1) ^k i^2 = (k+(1/3))((k(k+1))/2)−((k(k+1)(2k+1))/6) = ((k^2 (k+1))/6) (1/(k^2 (k+1)))=((−k+1)/k^2 )+(1/(k+1))= (1/k^2 )+(1/(k+1))−(1/k) Σ_(k=1) ^n (−(1/k)+(1/(k+1)))=−1+(1/2)+((−1)/2)+(1/3)+...+((−1)/n)+(1/(n+1))= −1+(1/(n+1))→_(n→∞) −1 lim_(n→∞) Σ_(k=1) ^n (1/k^2 )=(π^2 /6) Ω=6((π^2 /6)−1)=π^2 −6

$$ \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}{i}\left({k}−{i}+\frac{\mathrm{1}}{\mathrm{3}}\right)=\left({k}+\frac{\mathrm{1}}{\mathrm{3}}\right)\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}{i}−\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}{i}^{\mathrm{2}} = \\ $$$$\left({k}+\frac{\mathrm{1}}{\mathrm{3}}\right)\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}}−\frac{{k}\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)}{\mathrm{6}}\:= \\ $$$$\frac{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\frac{\mathrm{1}}{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)}=\frac{−{k}+\mathrm{1}}{{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{{k}+\mathrm{1}}= \\ $$$$\frac{\mathrm{1}}{{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{{k}+\mathrm{1}}−\frac{\mathrm{1}}{{k}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\frac{\mathrm{1}}{{k}}+\frac{\mathrm{1}}{{k}+\mathrm{1}}\right)=−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{−\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{−\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}= \\ $$$$−\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\underset{{n}\rightarrow\infty} {\rightarrow}−\mathrm{1} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Omega=\mathrm{6}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}\right)=\pi^{\mathrm{2}} −\mathrm{6} \\ $$

Commented by hardmath last updated on 09/Apr/25

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$

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