Studying the convergence of two sequences

Question

Let $(a _n)_{n\geq1}$,$\ (b _n)_{n\geq 1}$, so that $\ a_n={\ \log_{n}}\sqrt{\ n^2+n-1}$ and $b_n={\ \log_{n}}\sqrt{\ n^2+n+1}$. Prove they are convergent and find their limit. I have little ideas on how to accomplish this. Firstly, ${\ a_n}$ and ${\ b_n}$ are both sequences bigger than 1 and look alike. I also observed they don`t seem to get very far from $1$, but that could be false as I have no proof. Could you help me prove this?

Answer 1

$$\ a_n={\ \log_{n}}\sqrt{\ n^2+n-1}=$$

$$\log_{n} n\sqrt{\ 1+ \frac {1}{n}-\frac {1}{n^2}}=$$

$$\log_{n} n +\log_{n} \sqrt{\ 1+ \frac {1}{n}-\frac {1}{n^2}}=$$

$$ 1+\log_{n} \sqrt{\ 1+ \frac {1}{n}-\frac {1}{n^2}}\to 1$$ Similarly $$b_n={\ \log_{n}}\sqrt{\ n^2+n+1} \to 1$$

Answer 2

Notice that you have $$a_n = \log_n \sqrt{n^2+n-1} = \frac{\log(\sqrt{n^2+n-1})}{\log(n)} = \frac{\log(n^2+n-1)}{2\log(n)}$$ by logarithm laws. Now you can use L'Hopital's Rule to deduce $$ \lim_{n \to \infty} \frac{\log(n^2+n-1)}{2\log(n)} = \lim_{n \to \infty} \frac{\frac{2n + 1}{n^2+n-1}}{2 \frac 1 n } = \lim_{n \to \infty} \frac{2n^2 + n}{2n^2 + 2n - 2} = \lim_{n \to \infty} \frac{2 + \frac 1 n}{2 + \frac 2 n - \frac{2}{n^2}} = \frac 2 2 = 1.$$ The second sequence can be treated accordingly. You should try it yourself :)