Let $$\alpha = \lim_{n\to\infty}\sum_{m = n^2}^{2n^2}\dfrac{1}{\sqrt{5n^4 + n^3 + m}}.$$ Then what is $10\sqrt5\,\alpha$ is equal to?
I am trying to solve this problem as the limit of Riemann sum and I got stuck in the middle and I am unable to proceed further. Is my approach wrong or is there some other way to solve this?
On
For this kind of problems, generalized haronic numbers are very useful$$a_n=\sum_{m=n^2}^{2n^2}\frac{1}{\sqrt{5 n^4+n^3+m}}=H_{5 n^4+n^3+2 n^2}^{\left(\frac{1}{2}\right)}-H_{5 n^4+n^3+n^2-1}^{\left(\frac{1}{2}\right)}$$ Now, using $$H_p^{\left(\frac{1}{2}\right)}=2 \sqrt{p}+\zeta \left(\frac{1}{2}\right)+\frac 1{2\sqrt p}+O\left(\frac{1}{p^{3/2}}\right)$$ use it twice and continue with binomial expansion or Taylor series to get $$a_n=\frac 1{\sqrt 5} \left(1-\frac{1}{10 n}+\frac{173}{200 n^2}+O\left(\frac{1}{n^3}\right)\right)$$
I think you should not go into Riemann sums here, the most prudent approach seems to be the squeeze theorem. For the upper limit of the squeeze, in every term, let $m$ be equal to $n^2$ and for the lower limit, in every term let m be equal to $2n^2$. Both yield the same value, $1/\sqrt5$. So the answer should come out to be $10$.