Find a definite integral which represent $ \lim_{n\to\infty}{\frac{1}{n}\sum_{k=1}^{n}{\sqrt\frac{k}{n+k}}} $

Question

Find a definite integral which represent $$\lim\limits_{n\to\infty}{\frac{1}{n}\sum\limits_{k=1}^{n}{\sqrt\frac{k}{n+k}}}.$$ I don't know how can I approach the question. Is the answer $$\int_{0}^{1}{\sqrt\frac{x}{x+1}}dx\;?$$

Answer 1

You can write it simply \begin{align*}\lim\limits_{n\to\infty}{\frac{1}{n}\sum\limits_{k=1}^{n}{\sqrt\frac{k}{n+k}}} & = \lim\limits_{n\to\infty}{\frac{1}{n}\sum\limits_{k=1}^{n}{\sqrt\frac{\frac kn}{1+\frac kn}}} \\ & = \int_0^1\sqrt{\frac x{1+x}}dx. \end{align*} Or, with more explanation, if the Riemann integral $\int_0^1 f(x)\,dx$ exists, then it can be written as the limit

$$\int_a^b f(x)\,dx=\lim_{n\to \infty}\sum_{k=1}^n f\left(a+\frac{b-a}{n}\,k\right)\,\left(\frac{b-a}{n}\right).$$

Using the above formula with $f(x)=\sqrt{\frac x{1+x}}$, $a=0$ and $b=1$ reveals that $$\int_0^1\sqrt{\frac x{1+x}}dx=\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}{\sqrt\frac{\frac kn}{1+\frac kn}}{\frac{1}{n}}=\lim\limits_{n\to\infty}{\frac{1}{n}\sum\limits_{k=1}^{n}{\sqrt\frac{k}{n+k}}}.$$

Therefore, the limit of interest is simply the Riemann Sum of the integral $ \int_0^1\sqrt{\frac x{1+x}}dx$ .

Answer 2

$$\lim\limits_{n\to\infty}{\frac{1}{n}\sum\limits_{k=1}^{n}{\sqrt\frac{k}{n+k}}}$$ $$=\lim\limits_{n\to\infty}{\frac{1}{n}\sum\limits_{k=1}^{n}{\sqrt\frac{\frac{k}{n}}{1+\frac{k}{n}}}}$$ when we sub $$x=\frac{k}{n}$$ as sum k from 1 to n sum k from 1 to n means $\frac kn$ sum from $\frac 1n$ to 1

So we have $$\int_{0}^{1}\sqrt{\frac{x}{x+1}}dx$$