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Question Number 124738 by mnjuly1970 last updated on 05/Dec/20
             ...nice   calculus...   simple limit::     lim_(n→∞)  {((1^(a+1) +2^(a+1) +...+n^(a+1) )/(n(1^a +2^a +....n^a )))}=?   where a ≠−2 , −1
$$\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:\:{calculus}… \\ $$$$\:{simple}\:{limit}:: \\ $$$$\:\:\:{lim}_{{n}\rightarrow\infty} \:\left\{\frac{\mathrm{1}^{{a}+\mathrm{1}} +\mathrm{2}^{{a}+\mathrm{1}} +…+{n}^{{a}+\mathrm{1}} }{{n}\left(\mathrm{1}^{{a}} +\mathrm{2}^{{a}} +….{n}^{{a}} \right)}\right\}=? \\ $$$$\:{where}\:{a}\:\neq−\mathrm{2}\:,\:−\mathrm{1} \\ $$
Commented by PRITHWISH SEN 2 last updated on 07/Dec/20
lim_(n→∞)  ((Σ_(r=1) ^n ((r/n))^(a+1) )/(Σ_(k=1) ^n ((k/n))^a )) =((∫_0 ^1 x^(a+1) dx)/(∫_0 ^1 x^a dx)) = ((a+1)/(a+2))
$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\frac{\mathrm{r}}{\mathrm{n}}\right)^{\mathrm{a}+\mathrm{1}} }{\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\frac{\mathrm{k}}{\mathrm{n}}\right)^{\mathrm{a}} }\:=\frac{\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{a}+\mathrm{1}} \mathrm{dx}}{\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{a}} \mathrm{dx}}\:=\:\frac{\mathrm{a}+\mathrm{1}}{\mathrm{a}+\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 14/Dec/20
thank you sir..
$${thank}\:{you}\:{sir}.. \\ $$
Commented by PRITHWISH SEN 2 last updated on 14/Dec/20
welcome
$$\mathrm{welcome} \\ $$

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