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

Question

Among

(S1): $$\lim_\limits{n \rightarrow \infty} \frac{1}{n^{2}}(2+4+6+\ldots \ldots+2 n)=1$$

(S2) : $$\lim_\limits{n \rightarrow \infty} \frac{1}{n^{16}}\left(1^{15}+2^{15}+3^{15}+\ldots \ldots+n^{15}\right)=\frac{1}{16}$$

Accepted Answer

$$
\begin{aligned}
& S_1: \lim _{n \rightarrow \infty} \frac{1}{n^2}[2+4+6+\ldots+2 n] \\
& \lim _{n \rightarrow \infty} 2 \frac{n(n+1)}{2 n^2}=1 \\

& S_2: \lim _{n \rightarrow \infty} \frac{1}{n^{16}}\left(\sum r^{15}\right)=\lim _{n \rightarrow \infty} \frac{1}{n} \sum\left(\frac{r}{n}\right)^{15} \\
& =\int_0^1 x^{15} d x=\frac{1}{16} \\

& \therefore \text { Both } S_1 \text { and } S_2 \text { are correct. }
\end{aligned}
$$

Solutions · 1