Find: $\lim_{n \rightarrow \infty} \sum_{k = 1}^n \frac{\sqrt{\binom{n+k}{2}}}{n^2} $.
I am clueless how to solve this as I am unable to sandwich the above summation by two series which converge to the same limit (and hence use sandwich theorem). Neither do I know of any asymptotic approximation for the above summation, nor can I write the above summation as the Riemann sum of an integral. How do I proceed?
Edit: $\binom{n}{r}$ is the the binomial coefficient $^nC_r$.
On
$$\frac{1}{\sqrt{2}}\int_0^11+x dx =\frac{1}{\sqrt{2}n}\sum_{i=1}^n1+\frac{i-1}{n}\leq \frac{1}{\sqrt{2}n}\sum_{i=1}^n\sqrt{(1+\frac{i-1}{n})(1+\frac{i}{n}) }\leq \frac{1}{\sqrt{2}n}\sum_{i=1}^n1+\frac{i}{n} = \frac{1}{\sqrt{2}}\int_0^11+x dx $$
so the limit is $\frac{1}{\sqrt{2}}\int_0^11+x dx =\frac{3}{2\sqrt{2}}$.
As $\binom{n+k}{2}=\frac12(n+k)(n+k-1)$, in the limit we can use the fact that $\binom{n+k}{2}\approx\frac12(n+k)^2\implies\sqrt{\binom{n+k}{2}}\approx\frac{\sqrt2}2(n+k)$ $$\lim_{n \to\infty} \sum_{k = 1}^n \frac{\sqrt{\binom{n+k}{2}}}{n^2}=\lim_{n \to\infty} \frac{\sqrt2}2\sum_{k = 1}^n \frac{n+k}{n^2}$$ $$=\lim_{n \to\infty} \frac{\sqrt2}2\bigg(\sum_{k = 1}^n \frac{1}{n}+\frac{k}{n^2}\bigg)$$ $$=\lim_{n \to\infty} \frac{\sqrt2}2\bigg(\frac1n(n)+\frac{1}{n^2}\Big(\frac12n(n+1)\Big)\bigg)$$ $$=\lim_{n \to\infty} \frac{\sqrt2}2\bigg(1+\frac12(1+\frac1n)\bigg)$$ $$=\frac{\sqrt2}2\bigg(1+\frac12\bigg)=\frac{3\sqrt2}{4}$$