calculus - Finding a limit with squeeze theorem

Let $a_n={\frac1{\sqrt{n^2+1}}}+{\frac1{\sqrt{n^2+2}}}_{....}{\frac1{\sqrt{n^2+n+1}}} $

Find $\lim_{n\to\infty}a_n$.

I know the limit is zero. I also know that I need to use the squeeze theorem to solve it.

Now using ${\frac n{\sqrt{n^2+1}}}\le a_n \le {\frac n{\sqrt{n^2+n+1}}}$ as my upper and lower bounds doesn't yield 0, it yields 1.

It doesn't seem right but can I just use ${\frac 1{\sqrt{n^2+1}}}\le a_n \le {\frac 1{\sqrt{n^2+n+1}}}$ to solve ?

Thanks.

Answer

The limit is not 0, in fact you are almost there!

${\frac {n+1}{\sqrt{n^2+n+1}}}\le a_n \le {\frac {n+1}{\sqrt{n^2+1}}}$

For LHS, $\lim_{n\to\infty} \frac{n+1}{\sqrt{n^2+n+1}}=\lim_{n\to\infty}\frac{1+\frac{1}{n}}{\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}}=1$.

For RHS, $\lim_{n\to\infty} \frac{n+1}{\sqrt{n^2+1}}=\lim_{n\to\infty}\frac{1+\frac{1}{n}}{\sqrt{1+\frac{1}{n^2}}}=1$

Hence, by Squeeze Theorem, $\lim_{n\to\infty}a_n=1$