Lorena Lester

2022-07-22

Compute $\underset{n\to \mathrm{\infty }}{lim}\left(3n{\right)}^{1/3}{x}_{n}$

yelashwag8

Expert

We have that
${x}_{n}\to 0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{S}_{n}=\sum _{k=1}^{n}{x}_{k}^{2}\to \mathrm{\infty }$
We use Stolz's lemma to show
$\underset{n}{lim}\frac{3n}{{S}_{n}^{3}}=1$
And we would be done after that, indeed
$\underset{n}{lim}3n{x}_{n}^{3}=\underset{n}{lim}\frac{3n{x}_{n}^{3}}{{x}_{n}^{3}{S}_{n}^{3}}=\underset{n}{lim}\frac{3n}{{S}_{n}^{3}}=1$
and so by continuity of $f\left(x\right)={x}^{\frac{1}{3}}$ we are done.
Now let's prove our claim
$\underset{n}{lim}\frac{3n}{{S}_{n}^{3}}=\underset{n}{lim}\frac{3\left(n+1\right)-3n}{{S}_{n+1}^{3}-{S}_{n}^{3}}=$
$=\frac{3}{\left({S}_{n+1}-{S}_{n}\right)\left({S}_{n+1}^{2}+{S}_{n+1}{S}_{n}+{S}_{n}^{2}\right)}=$
$=\frac{3}{{x}_{n+1}^{2}\left({S}_{n+1}^{2}+{S}_{n+1}{S}_{n}+{S}_{n}^{2}\right)}=$
$=\frac{3}{{x}_{n+1}^{2}{S}_{n+1}^{2}\left(1+\frac{{S}_{n}}{{S}_{n+1}}+\frac{{S}_{n}}{{S}_{n+1}}{\right)}^{2}}=$
Now
$\underset{n}{lim}\frac{{S}_{n}}{{S}_{n+1}}=\underset{n}{lim}\left(1-\frac{{x}_{n+1}^{2}}{{S}_{n+1}}\right)=1$
Because
$\frac{{x}_{n}^{2}}{{S}_{n}}=\frac{{x}_{n}^{3}}{{S}_{n}{x}_{n}}$
So we are left
$\underset{n}{lim}\frac{3n}{{S}_{n}^{3}}=\underset{n}{lim}\frac{3}{{x}_{n+1}^{2}{S}_{n+1}^{2}}\frac{1}{3}=1$

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