Cómo calcular el límite $\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n$

Question

<span class="math-container">$$I=\lim{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n$$</span> Probé <span class="math-container">$$\frac{n}{\sqrt{n^2+n+n}}\leq\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\leq\frac{n}{\sqrt{n^2+n+1}}$$</span> Pero<span class="math-container">$$\lim{n\to\infty}\left(\frac{n}{\sqrt{n^2+n+n}}\right)^n=\lim{n\to\infty}\left(1+\frac2n\right)^{\displaystyle-\frac{n}{2}}=e^{\displaystyle \lim{n\to\infty}-\frac n2\ln(1+\frac2n)}=\frac1e$$</span> Y de la misma manera, obtuve <span class="math-container">$$\lim{n\to\infty}\left(\frac{n}{\sqrt{n^2+n+1}}\right)^n=\lim{n\to\infty}\left(1+\frac1n+\frac{1}{n^2}\right)^{\displaystyle-\frac{n}{2}}=e^{\displaystyle \lim_{n\to\infty}-\frac n2\ln(1+\frac1n+\frac{1}{n^2})}=\frac{1}{\sqrt e}$$</span> Así que sólo tengo <span class="math-container">$$\frac1e\leq I\leq\frac{1}{\sqrt e}$$</span> ¿Podría alguien ayudarme a obtener el valor de <span class="math-container">$I$</span>. ¡gracias!

Answer 1

El registro natural de esta expresión es <span class="math-container">$$ K= n\ln( 1+ (x-1) )$$ where <span class="math-container">$$x=\sum_{r=1}^n \frac{1}{\sqrt{n^2+n+r}} $$</span> Due to the bounds you have shown, <span class="math-container">$K$</span> must tend to a finite non-zero number, which can only happen if <span class="math-container">$x\to 1$</span>. Now, <span class="math-container">$$K = n(x-1) +n O((x-1)^2)$$</span> The first term is <span class="math-container">$$n \sum_1^n \left( \frac{1}{\sqrt{n^2+n+r}} -\frac 1n \right) \ = \sum_1^n \left( \frac{1}{\sqrt{ 1+\frac 1n +\frac{r}{n^2}}} -1 \right) \ = \sum_1^n \left( -\frac 12 \left( \frac 1n +\frac{r}{n^2} \right) +O(\frac{1}{n^2}) \right) \ = -\frac 12 -\frac{n(n+1)}{4n^2} +O(\frac 1n ) \ \to -\frac 12 -\frac 14 \ =-\frac 34$$</span> The rest of the terms in <span class="math-container">$K$</span> can be shown to go to zero in a similar way. Hence, your limit is <span class="math-container">$$\Large{ \color{blue}{e^{-\frac 34 }} }$%$$</span></span>