Among (S1): n lim1n^2(2+4+6+ +2n)=1 (S2) : n lim1n^16(1^15+2^15+3^15+ +n^15)=(1)/(16)

Question

Among

(S1): $\underset{n	o \infty}{lim}\frac{1}{{n}^{2}}(2+4+6+\ldots \ldots +2n)=1$

(S2) : $\underset{n	o \infty}{lim}\frac{1}{{n}^{16}}({1}^{15}+{2}^{15}+{3}^{15}+\ldots \ldots +{n}^{15})=\frac{1}{16}$

Accepted Answer

$\begin{gathered} {S}_{1}:\underset{n	o \infty}{lim}\frac{1}{{n}^{2}}[2+4+6+\ldots +2n] \ \underset{n	o \infty}{lim}2\frac{n(n+1)}{2{n}^{2}}=1 \ {S}_{2}:\underset{n	o \infty}{lim}\frac{1}{{n}^{16}}(\sum {r}^{15})=\underset{n	o \infty}{lim}\frac{1}{n}\sum {(\frac{r}{n})}^{15} \ ={\int }_{1}^{0}{x}^{15}dx=\frac{1}{16} \ 	herefore {S}_{1}{S}_{2} \end{gathered}$

Solutions · 1