Evaluate limit using definite integral

Question

I've got this, but i think i took a wrong turn and i don't know what to do next $$\lim_{n\to\infty} \frac{\sqrt{n^2 + 1} + \ldots + \sqrt{n^2 + n^2}}{n^2} = ?$$ $$\frac{\sqrt{n^2 + 1} + \ldots + \sqrt{n^2 + n^2}}{n^2}=\sum_{k=1}^{n^2} \frac{\sqrt{n^2 + k}}{n^2} = \sum_{k=1}^{n^2} \frac{\sqrt{n^2 + k}}{n^2} = \sum_{k=1}^{n^2} \frac{\sqrt{n^2 + k}}{n} \cdot \frac{1}{n} = \sum_{k=1}^{n^2} \sqrt{1 + \frac{k}{n^2}} \cdot \frac{1}{n}$$

Answer 1

Hint:

I think you should rewrite the sum as $$\frac1n\sum_{k=1}^{n} \frac{\sqrt{n^2 + k^2}}{n}= \frac1n\sum_{k=1}^{n} \sqrt{\frac{n^2 + k^2\strut}{n^2}}=\frac1n\sum_{k=1}^{n} \sqrt{1+\frac{k^2\mathstrut}{n^2}\,.}$$

Answer 2

If we assume $$\sqrt{n^2 + 1} + \cdots + \sqrt{n^2 + n^2}$$ was intended to mean the sum $$ \sqrt{n^2 + 1} + \sqrt{n^2 + 2} + \sqrt{n^2 + 3} + \cdots + \sqrt{n^2 + n^2} $$ of $n^2$ terms, then \begin{align*} &\frac{\sqrt{n^2 + 1} + \cdots + \sqrt{n^2 + n^2}}{n^2}\\[4pt] =\;&\sum_{k=1}^{n^2} \frac{\sqrt{n^2 + k}}{n^2}\\[2pt] >\;&\sum_{k=1}^{n^2} \frac{\sqrt{n^2}}{n^2}\\[2pt] =\;&\sum_{k=1}^{n^2} \frac{1}{n}\\[2pt] =\;&n^2\cdot \frac{1}{n}\\[2pt] =\;&n\\[4pt] \end{align*} hence $$\lim_{n\to\infty} \frac{\sqrt{n^2 + 1} + \cdots + \sqrt{n^2 + n^2}}{n^2} = \infty$$

Alternatively, applying the concept of Riemann sums, \begin{align*} &\lim_{n\to\infty} \frac{\sqrt{n^2 + 1} + \cdots + \sqrt{n^2 + n^2}}{n^3}\\[4pt] =\;&\lim_{n\to\infty} \;\sum_{k=1}^{n^2} \frac{\sqrt{n^2 + k}}{n^3}\\[2pt] =\;&\lim_{n\to\infty} \;\sum_{k=1}^{n^2} \frac{\sqrt{n^2 + k}}{n}\cdot\frac{1}{n^2}\\[2pt] =\;&\lim_{n\to\infty} \;\sum_{k=1}^{n^2} \sqrt{1 + \frac{k}{n^2}}\cdot\frac{1}{n^2}\\[2pt] =\;&\int_0^1\!\sqrt{1+x}\;dx\\[2pt] =\;&c,\;\text{where}\;c\in(0,\infty)\\[2pt] \end{align*} In fact, the actual value of $c$ is ${\large{\frac{4\sqrt{2}-2}{3}}}$, but for the argument that follows, we only need $c\in(0,\infty)$.

So now we have \begin{align*} &\lim_{n\to\infty} \frac{\sqrt{n^2 + 1} + \cdots + \sqrt{n^2 + n^2}}{n^2}\\[4pt] =\;&\lim_{n\to\infty} \frac{\sqrt{n^2 + 1} + \cdots + \sqrt{n^2 + n^2}}{n^3}\cdot n\\[2pt] =\;&\lim_{n\to\infty} c\cdot n\\[2pt] =\;&c\cdot\!\lim_{n\to\infty} n\\[2pt] =\;&c\cdot\infty\\[2pt] =\;&\infty\\[2pt] \end{align*} However, noting the comments by Upstart, ncmathsadist, and robjohn, and also Bernard's answer, I suspect that $$\sqrt{n^2 + 1} + \cdots + \sqrt{n^2 + n^2}$$ was actually intended to mean $$ \sqrt{n^2 + 1^2} + \sqrt{n^2 + 2^2} + \sqrt{n^2 + 3^2} + \cdots + \sqrt{n^2 + n^2} $$ which is a sum of $n$ terms.

With that assumption, \begin{align*} &\lim_{n\to\infty} \frac{\sqrt{n^2 + 1^2} + \cdots + \sqrt{n^2 + n^2}}{n^2}\\[6pt] =\;&\lim_{n\to\infty} \;\sum_{k=1}^n \frac{\sqrt{n^2 + k^2}}{n^2}\\[6pt] =\;&\lim_{n\to\infty} \;\sum_{k=1}^n \frac{\sqrt{n^2 + k^2}}{n}\cdot\frac{1}{n}\\[6pt] =\;&\lim_{n\to\infty} \;\sum_{k=1}^n \sqrt{1 + \left({\small{\frac{k}{n}}}\right)^2}\cdot\frac{1}{n}\\[4pt] =\;&\int_0^1\!\sqrt{1+x^2}\;dx\\[4pt] =\;&\frac{\sqrt{2}-\ln\left(\sqrt{2}-1\right)}{2}\\[4pt] \end{align*}

Answer 3

Too long for a comment.

Without Riemann sums, we could have a quite good approximation using generalized harmonic numbers since $$S_n=\sum_{i=1}^{n^2} \sqrt{n^2+i}=H_{2 n^2}^{\left(-\frac{1}{2}\right)}-H_{n^2}^{\left(-\frac{1}{2}\right)}$$ Now, using the asymptotics $$H_{p}^{\left(-\frac{1}{2}\right)}=\frac{2 p^{3/2}}{3}+\frac{\sqrt{p}}{2}+\zeta \left(-\frac{1}{2}\right)+\frac{1}{24\sqrt{{p}}}+O\left(\frac{1}{p^{5/2}}\right)$$ $$S_n=\frac{2}{3} \left(2 \sqrt{2}-1\right) n^3+\left(\frac{1}{\sqrt{2}}-\frac{1}{2}\right) n+\frac{\sqrt{2}-2}{48 n}+O\left(\frac{1}{n^5}\right)$$ $$\frac{S_n}{n^2}=\frac{2}{3} \left(2 \sqrt{2}-1\right) n+\frac{\frac{1}{\sqrt{2}}-\frac{1}{2}}{n}+\frac{\sqrt{2}-2}{48 n^3}+O\left(\frac{1}{n^7}\right)$$

Uisng the above approximation, for $n=10$ we would obtain $\frac{642401 \sqrt{2}-322402}{48000}$ which is $\approx 12.210212639209$ while the exact summation would lead to $\approx 12.210212639252$