Menu Close

n-0-1-n-1-n-n-

Tinku Tara 0 Comments
Pin




Question Number 205774 by mnjuly1970 last updated on 30/Mar/24
−−−−−−− Ω = Σ_(n=0) ^∞ (( (−1)^( n) )/((−1)^( n) −n)) = ? −−−−−−−
$$ \\ $$$$\:\:\:\:\:\:−−−−−−− \\ $$$$\:\:\:\:\:\Omega\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{\:{n}} }{\left(−\mathrm{1}\right)^{\:{n}} \:−{n}}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:−−−−−−− \\ $$
Answered by MathedUp last updated on 30/Mar/24
Ω=Σ (((−1)^h )/((−1)^h −h))=Σ ((i^(2h) (i^(2h) +h))/((i^(2h) −h)(i^(2h) +h)))=Σ ((i^(4h) +h∙i^(2h) )/((i^(4h) −h^2 ))) Σ ((1+h∙(−1)^h )/(1−h^2 ))=1+(1/2)+Σ_(h=2) ^∞ ((1+h(−1)^h )/(1−h^2 )) a_0 =1 a_1 =−lim_(h→1) ((1+he^(iπh) )/(h^2 −1))=−lim_(h→1) ((e^(iπh) +iπhe^(iπh) )/(2h)) −((e^(iπ) +iπe^(iπ) )/2)=(1/2)+(1/2)iπ Re{a_1 }=(1/2) (3/2)−Σ_(h=2) ^∞ ((1+he^(iπh) )/(h^2 −1))= approximatly 0.3068678195....
$$\Omega=\Sigma\:\frac{\left(−\mathrm{1}\right)^{{h}} }{\left(−\mathrm{1}\right)^{{h}} −{h}}=\Sigma\:\frac{\boldsymbol{{i}}^{\mathrm{2}{h}} \left(\boldsymbol{{i}}^{\mathrm{2}{h}} +{h}\right)}{\left(\boldsymbol{{i}}^{\mathrm{2}{h}} −{h}\right)\left(\boldsymbol{{i}}^{\mathrm{2}{h}} +{h}\right)}=\Sigma\:\frac{\boldsymbol{{i}}^{\mathrm{4}{h}} +{h}\centerdot\boldsymbol{{i}}^{\mathrm{2}{h}} }{\left(\boldsymbol{{i}}^{\mathrm{4}{h}} −{h}^{\mathrm{2}} \right)} \\ $$$$\Sigma\:\frac{\mathrm{1}+{h}\centerdot\left(−\mathrm{1}\right)^{{h}} }{\mathrm{1}−{h}^{\mathrm{2}} }=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\underset{{h}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}+{h}\left(−\mathrm{1}\right)^{{h}} }{\mathrm{1}−{h}^{\mathrm{2}} } \\ $$$${a}_{\mathrm{0}} =\mathrm{1}\:{a}_{\mathrm{1}} =−\underset{{h}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}+{he}^{\boldsymbol{{i}}\pi{h}} }{{h}^{\mathrm{2}} −\mathrm{1}}=−\underset{{h}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{e}^{\boldsymbol{{i}}\pi{h}} +\boldsymbol{{i}}\pi{he}^{\boldsymbol{{i}}\pi{h}} }{\mathrm{2}{h}} \\ $$$$−\frac{{e}^{\boldsymbol{{i}}\pi} +\boldsymbol{{i}}\pi{e}^{\boldsymbol{{i}}\pi} }{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{i}}\pi\:\:\mathrm{Re}\left\{{a}_{\mathrm{1}} \right\}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}−\underset{{h}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}+{he}^{\boldsymbol{{i}}\pi{h}} }{{h}^{\mathrm{2}} −\mathrm{1}}=\:{approximatly} \\ $$$$\mathrm{0}.\mathrm{3068678195}…. \\ $$
Answered by Berbere last updated on 30/Mar/24
Ω=Σ_(n=0) ^∞ ((1/(1−2n))−(1/(−2n−2))) =Σ_(n≥0) ((1/(1−2n))+(1/(2(n+1))))=Σ_(n≥0) ((1/(2(n+1)))−(1/(2n−1))) =(3/2)−Σ_(n≥1) (1/(2n+1+1))−(1/(2n−2+1)) =(3/2)+Σ_(n≥1) (∫_0 ^1 x^(2n+1) −x^(2n−2) dx) S(N)=Σ_(n≥0) ∫_0 ^1 (x^3 /(1−x^2 ))−(1/(1−x^2 ))=(3/2)+∫_0 ^1 ((x^3 −1)/((1−x)(1+x)))dx =(3/2)−∫_0 ^1 ((x^2 +x+1)/(x+1))=(3/2)−∫_0 ^1 ((x(x+1)+1)/(x+1))dx =(3/2)−∫_0 ^1 x+(1/(1+x))dx=1−ln(2) Ω=1−ln(2)
$$\Omega=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}{n}}−\frac{\mathrm{1}}{−\mathrm{2}{n}−\mathrm{2}}\right) \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{2}+\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} −{x}^{\mathrm{2}{n}−\mathrm{2}} {dx}\right) \\ $$$${S}\left({N}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{2}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{3}} −\mathrm{1}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}{dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{{x}+\mathrm{1}}=\frac{\mathrm{3}}{\mathrm{2}}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}\left({x}+\mathrm{1}\right)+\mathrm{1}}{{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}−\int_{\mathrm{0}} ^{\mathrm{1}} {x}+\frac{\mathrm{1}}{\mathrm{1}+{x}}{dx}=\mathrm{1}−{ln}\left(\mathrm{2}\right) \\ $$$$\Omega=\mathrm{1}−{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 30/Mar/24
so good solution
$${so}\:{good}\:{solution} \\ $$
Answered by MM42 last updated on 30/Mar/24
Ω=((1/1))+(((−1)/(−1−1)))+((1/(1−2)))+(((−1)/(−1−3)))+((1/(1−4)))+(((−1)/(−1−5)))+((1/(1−6)))+(((−1)/(−1−7)))... 1+(1/2)−1+(1/4)−(1/3)+(1/6)−(1/5)+(1/8)−... =1−(1−(1/2)+(1/3)−(1/4)+(1/5)−(1/6)+..) =1−ln2 ✓
$$\Omega=\left(\frac{\mathrm{1}}{\mathrm{1}}\right)+\left(\frac{−\mathrm{1}}{−\mathrm{1}−\mathrm{1}}\right)+\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}}\right)+\left(\frac{−\mathrm{1}}{−\mathrm{1}−\mathrm{3}}\right)+\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{4}}\right)+\left(\frac{−\mathrm{1}}{−\mathrm{1}−\mathrm{5}}\right)+\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{6}}\right)+\left(\frac{−\mathrm{1}}{\left.−\mathrm{1}−\mathrm{7}\right)}…\right. \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{8}}−… \\ $$$$=\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{6}}+..\right) \\ $$$$=\mathrm{1}−{ln}\mathrm{2}\:\checkmark \\ $$$$ \\ $$
Commented by MathedUp last updated on 30/Mar/24
good good good
$$\mathrm{good}\:\mathrm{good}\:\mathrm{good} \\ $$
Commented by mnjuly1970 last updated on 30/Mar/24
thanks alot
$${thanks}\:{alot} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *