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Question Number 196327 by pticantor last updated on 22/Aug/23
calcul la somme suivante: lim_(n→+∞) Σ_(k=n) ^(2n) sin((𝛑/k)) elrochi
$$\boldsymbol{{calcul}}\:\boldsymbol{{la}}\:\boldsymbol{{somme}}\:\boldsymbol{{suivante}}: \\ $$$$\:\:\boldsymbol{{li}}\underset{\boldsymbol{{n}}\rightarrow+\infty} {\boldsymbol{{m}}}\:\underset{\boldsymbol{{k}}=\boldsymbol{{n}}} {\overset{\mathrm{2}\boldsymbol{{n}}} {\sum}}\boldsymbol{{sin}}\left(\frac{\boldsymbol{\pi}}{\boldsymbol{{k}}}\right) \\ $$$$\:\:\boldsymbol{{elrochi}} \\ $$
Answered by sniper237 last updated on 22/Aug/23
for x>0 , 0<sinx<x ⇒ 1−(x^2 /2)<cosx<1 ⇒ x−(x^3 /6)< sinx<x. Then for x=π/k Σ_(k=n) ^(2n) (π/k) −Σ_(k=n) ^(2n) (π^3 /k^3 ) < Σ_(k=n) ^(2n) sin((π/k))<Σ_(k=n) ^(2n) (π/k) Σ_n ^(2n) (1/k^3 ) =Σ_(k=1) ^(2n) (1/k^3 )−Σ_(k=1) ^(n−1) (1/k^3 ) →_∞ ζ(3)−ζ(3)=0 Σ_(k=n) ^(2n) (1/k) =(1/n) Σ_(k=0) ^n (1/(1+(k/n))) →_∞ ∫_0 ^1 (dx/(1+x)) =ln2 gendarm theorem ⇒ Answer=πln2
$${for}\:{x}>\mathrm{0}\:,\:\:\:\mathrm{0}<{sinx}<{x}\:\Rightarrow\:\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}<{cosx}<\mathrm{1} \\ $$$$\Rightarrow\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}<\:{sinx}<{x}.\:{Then}\:{for}\:{x}=\pi/{k} \\ $$$$\:\:\:\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\pi}{{k}}\:−\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\pi^{\mathrm{3}} }{{k}^{\mathrm{3}} }\:<\:\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\sum}}{sin}\left(\frac{\pi}{{k}}\right)<\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\pi}{{k}} \\ $$$$\underset{{n}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\:=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{3}} }−\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\:\underset{\infty} {\rightarrow}\:\zeta\left(\mathrm{3}\right)−\zeta\left(\mathrm{3}\right)=\mathrm{0} \\ $$$$\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:=\frac{\mathrm{1}}{{n}}\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{1}+\frac{{k}}{{n}}}\:\underset{\infty} {\rightarrow}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}}\:={ln}\mathrm{2} \\ $$$$\:{gendarm}\:{theorem}\:\Rightarrow\:\:{Answer}=\pi{ln}\mathrm{2} \\ $$
Answered by witcher3 last updated on 22/Aug/23
sin(x)=x−(x^3 /6)+c(x^4 ) x−(x^3 /6)≤sin(x)≤x this solve Quation Σ_n ^(2n) (π/k)=π(H_(2n) −H_(n−1) )=π(H_(2n) −ln(2n)−H_(n−1) +ln(n−1)+ln(((2n)/(n−1)))) H_(ϕ(n)) −ln(ϕ(n))→γ, Euler Constante∀ϕ bijective N→^ϕ N for x^3 therme Σ_(k=n) ^(2n) (1/k^3 )<n.(1/n^3 )<(1/n^2 )→0...Try withe This Hint
$$\mathrm{sin}\left(\mathrm{x}\right)=\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}+\mathrm{c}\left(\mathrm{x}^{\mathrm{4}} \right) \\ $$$$\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\leqslant\mathrm{sin}\left(\mathrm{x}\right)\leqslant\mathrm{x} \\ $$$$\mathrm{this}\:\mathrm{solve}\:\mathrm{Quation} \\ $$$$\underset{\mathrm{n}} {\overset{\mathrm{2n}} {\sum}}\frac{\pi}{\mathrm{k}}=\pi\left(\mathrm{H}_{\mathrm{2n}} −\mathrm{H}_{\mathrm{n}−\mathrm{1}} \right)=\pi\left(\mathrm{H}_{\mathrm{2n}} −\mathrm{ln}\left(\mathrm{2n}\right)−\mathrm{H}_{\mathrm{n}−\mathrm{1}} +\mathrm{ln}\left(\mathrm{n}−\mathrm{1}\right)+\mathrm{ln}\left(\frac{\mathrm{2n}}{\mathrm{n}−\mathrm{1}}\right)\right) \\ $$$$\mathrm{H}_{\varphi\left(\mathrm{n}\right)} −\mathrm{ln}\left(\varphi\left(\mathrm{n}\right)\right)\rightarrow\gamma,\:\:\mathrm{Euler}\:\mathrm{Constante}\forall\varphi\:\mathrm{bijective}\:\mathbb{N}\overset{\varphi} {\rightarrow}\mathbb{N} \\ $$$$\mathrm{for}\:\mathrm{x}^{\mathrm{3}} \mathrm{therme} \\ $$$$\underset{\mathrm{k}=\mathrm{n}} {\overset{\mathrm{2n}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{3}} }<\mathrm{n}.\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }<\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\rightarrow\mathrm{0}…\mathrm{Try}\:\mathrm{withe}\:\mathrm{This}\:\mathrm{Hint} \\ $$$$ \\ $$$$ \\ $$

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