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L-lim-n-k-1-n-k-n-2-k-2-1-0-e-x-2-dx-1-m-0-1-m-2m-1-3-m-

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Question Number 220764 by Nicholas666 last updated on 18/May/25
L= lim _( n→∞) (Σ_(k=1) ^n (k/(n^2 +k^2 ))).(∫^( 1) _( 0) e^(−x^2 ) dx)^(−1) .(Σ_(m=0) ^∞ (((−1)^m )/((2m+1)3^m )))
$$ \\ $$$$\:\:\boldsymbol{\mathrm{L}}=\:\boldsymbol{\mathrm{lim}}\underset{\:\boldsymbol{{n}}\rightarrow\infty} {\:}\left(\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\sum}}\:\frac{\boldsymbol{{k}}}{\boldsymbol{{n}}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} }\right).\left(\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \boldsymbol{{e}}^{−\boldsymbol{{x}}^{\mathrm{2}} } \boldsymbol{{dx}}\overset{−\mathrm{1}} {\right)}.\left(\underset{\boldsymbol{{m}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{m}}} }{\left(\mathrm{2}\boldsymbol{{m}}+\mathrm{1}\right)\mathrm{3}^{\boldsymbol{{m}}} }\right)\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$
Answered by breniam last updated on 19/May/25
lim_(n→∞) Σ_(k=1) ^n (k/(n^2 +k^2 ))=lim_(n→∞) (1/n)Σ_(k=1) ^n ((((k/n)))/(1+((k/n))^2 ))=∫_0 ^1 (x/(x^2 +1))dx=ln(√2) Σ_(m=0) ^∞ (((−1)^m )/((2m+1)3^m ))=∫_0 ^1 Σ_(m=0) ^∞ (−(x^2 /3))^m dx=∫_0 ^1 (1/((x^2 /3)+1))dx={x^− =(x/( (√3)))}=(√3)∫_0 ^(1/( (√3))) (1/(x^2 +1))dx=(((√3)π)/6) (1/(∫_0 ^1 e^(−x^2 ) dx))=((√π)/( 2erf(1)))
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}}{{n}^{\mathrm{2}} +{k}^{\mathrm{2}} }=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(\frac{{k}}{{n}}\right)}{\mathrm{1}+\left(\frac{{k}}{{n}}\right)^{\mathrm{2}} }=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}=\mathrm{ln}\sqrt{\mathrm{2}} \\ $$$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{\left(\mathrm{2}{m}+\mathrm{1}\right)\mathrm{3}^{{m}} }=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\right)^{{m}} \mathrm{d}{x}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{\frac{{x}^{\mathrm{2}} }{\mathrm{3}}+\mathrm{1}}\mathrm{d}{x}=\left\{\overset{−} {{x}}=\frac{{x}}{\:\sqrt{\mathrm{3}}}\right\}=\sqrt{\mathrm{3}}\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} {\int}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}=\frac{\sqrt{\mathrm{3}}\pi}{\mathrm{6}} \\ $$$$\frac{\mathrm{1}}{\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{e}^{−{x}^{\mathrm{2}} } \mathrm{d}{x}}=\frac{\sqrt{\pi}}{\:\mathrm{2erf}\left(\mathrm{1}\right)} \\ $$

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