I want to evaluate the following limit equation. $$ \lim_{n\to\infty}\left({1\over\sqrt{n^2+1^2}}+{1\over\sqrt{n^2+2^2}}+\dots+{1\over\sqrt{n^2+n^2}}\right) $$
$$\begin{align} \lim_{n\to\infty}\left({1\over\sqrt{n^2+1^2}}+{1\over\sqrt{n^2+2^2}}+\dots+{1\over\sqrt{n^2+n^2}}\right)&=\lim_{n\to\infty}\sum_{i=1}^{n}{1\over\sqrt{n^2+i^2}} \end{align}$$
My tries
I assume that $~n~$ takes a natural number.
$$ \lim_{n\to\infty}\sum_{i=1}^{n}{1\over\sqrt{n^2+n^2}}\leq\lim_{n\to\infty}\sum_{i=1}^{n}{1\over\sqrt{n^2+i^2}}\leq\lim_{n\to\infty}\sum_{i=1}^{n}{1\over\sqrt{n^2+1}} $$
$$ \lim_{n\to\infty}{n\over\sqrt{n^2+n^2}}\leq\lim_{n\to\infty}\sum_{i=1}^{n}{1\over\sqrt{n^2+i^2}}\leq\lim_{n\to\infty}{n\over\sqrt{n^2+1}} $$
$$ \lim_{n\to\infty}{n\over\sqrt{2n^2}}\leq\lim_{n\to\infty}\sum_{i=1}^{n}{1\over\sqrt{n^2+i^2}}\leq\lim_{n\to\infty}{n\over\sqrt{n^2\left(1+{1\over n^2}\right)}} $$
$$ \lim_{n\to\infty}{n\over\sqrt{2}n}\leq\lim_{n\to\infty}\sum_{i=1}^{n}{1\over\sqrt{n^2+i^2}}\leq\lim_{n\to\infty}{n\over n\sqrt{\left(1+{1\over n^2}\right)}} $$
$$ \lim_{n\to\infty}{1\over\sqrt{2}}\leq\lim_{n\to\infty}\sum_{i=1}^{n}{1\over\sqrt{n^2+i^2}}\leq\lim_{n\to\infty}{1\over \sqrt{\left(1+{1\over n^2}\right)}} $$
$$ \color{fuchsia}{0.707<{1\over\sqrt{2}}\leq\lim_{n\to\infty}\sum_{i=1}^{n}{1\over\sqrt{n^2+i^2}}\leq1 } $$
Can I proceed from here? Which means that this math problem is quoted from the college transfer exam in 1998 so I am worrying about that whether this pink inequalities got subtracted of score in an exam, as a final answer.
The sum can be manipulated into a Riemann sum
$$\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{\sqrt{n^2+k^2}} = \lim_{n\to\infty}\sum_{k=1}^n\frac{1}{\sqrt{1+\frac{k^2}{n^2}}}\cdot\frac{1}{n}\equiv \lim_{n\to\infty}\sum_{k=1}^nf(x_k)\Delta x$$
which converges to the integral
$$\longrightarrow \int_0^1\frac{dx}{\sqrt{1+x^2}} = \sinh^{-1}(1)$$