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$\lim _{n \rightarrow \infty} \frac{\left(1^{2}+2^{2}+\ldots+n^{2}\right) \sqrt[n]{n}}{(n+1)(n+10)(n+100)}=$
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Verified Answer
The correct answer is:
$\frac{1}{3}$
We have,
$$
\begin{aligned}
\lim _{n \rightarrow \infty} \frac{\left(1^{2}+2^{2}+\ldots+n^{2}\right) \sqrt[n]{n}}{(n+1)(n+10)(n+100)} \\
=& \lim _{n \rightarrow \infty} \frac{n(n+1)(2 n+n)}{6(n+1)(n+10)(n+100)}\left(\lim _{n \rightarrow \infty} \sqrt[n]{n}\right) \\
=& \lim _{n \rightarrow \infty} \frac{2 n^{2}+n}{6(n+10)(n+100)} \\
=& \lim _{n \rightarrow \infty} \frac{2+\frac{1}{n}}{6\left(1+\frac{10}{n}\right)\left(1+\frac{100}{n}\right)} \\
=& \frac{2+0}{6(1+0)(1+0)} \\
=& \frac{2}{6}=\frac{1}{3}
\end{aligned}
$$
$$
\begin{aligned}
\lim _{n \rightarrow \infty} \frac{\left(1^{2}+2^{2}+\ldots+n^{2}\right) \sqrt[n]{n}}{(n+1)(n+10)(n+100)} \\
=& \lim _{n \rightarrow \infty} \frac{n(n+1)(2 n+n)}{6(n+1)(n+10)(n+100)}\left(\lim _{n \rightarrow \infty} \sqrt[n]{n}\right) \\
=& \lim _{n \rightarrow \infty} \frac{2 n^{2}+n}{6(n+10)(n+100)} \\
=& \lim _{n \rightarrow \infty} \frac{2+\frac{1}{n}}{6\left(1+\frac{10}{n}\right)\left(1+\frac{100}{n}\right)} \\
=& \frac{2+0}{6(1+0)(1+0)} \\
=& \frac{2}{6}=\frac{1}{3}
\end{aligned}
$$
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