Question Number 128065 by 0731619177 last updated on 04/Jan/21
Commented by mnjuly1970 last updated on 04/Jan/21
![lim_(n→∞ ) Σ_(k=1 ) ^∞ (1/k^2 )=(π^2 /6) lim_(n→∞) (Σ_(k=1) ^n (1/k) −ln(n))=γ (euler− constant) lim_(x→∞) ((((x!))^(1/x) /x))=(1/e) ∴ ans: [ ((1/e))^γ ]^(π^2 /6) =((1/e))^((γπ^2 )/6)](https://www.tinkutara.com/question/Q128067.png)
$${lim}_{{n}\rightarrow\infty\:\:\:} \underset{{k}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$${lim}_{{n}\rightarrow\infty} \left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:\:−{ln}\left({n}\right)\right)=\gamma\:\left({euler}−\:{constant}\right) \\ $$$${lim}_{{x}\rightarrow\infty} \left(\frac{\sqrt[{{x}}]{{x}!}}{{x}}\right)=\frac{\mathrm{1}}{{e}} \\ $$$$\therefore\:\:\:\:{ans}:\:\:\left[\:\left(\frac{\mathrm{1}}{{e}}\right)^{\gamma} \:\right]^{\frac{\pi^{\mathrm{2}} }{\mathrm{6}}} =\left(\frac{\mathrm{1}}{{e}}\right)^{\frac{\gamma\pi^{\mathrm{2}} }{\mathrm{6}}} \\ $$