# Why $\int_2^\infty{1\over{\sqrt x(\sqrt x-1)^2}}dx \style{text-decoration:line-through}{\approx} \sum_{n=2}^{\infty}{1\over{\sqrt n(\sqrt n-1)^2}}$?

I am aware that there are certain cases where the infinite sum does not equal the infinite integral. However, I am not yet advanced enough to be able to understand the Euler–Maclaurin formula, especially in terms of coming up with the $$k$$th Bernoulli Number. That said, I am wondering if there is a way somebody could explain to me the following inequality that came up in my course. When I use a calculator to estimate the infinite sum $$\sum_{n=2}^{\infty}{1\over{\sqrt n(\sqrt n-1)^2}}$$, I get $$\sum_{n=2}^{9999999}{1\over{\sqrt n(\sqrt n-1)^2}} \approx 7.47356$$

When I try to estimate using an integral, I get a huge discrepancy $$\int_2^\infty{1\over{\sqrt x(\sqrt x-1)^2}}dx=2^{3\over2}+2 \approx 4.8284$$

Does this require the Euler–Maclaurin formula to explain, or is there an easier way to understand what is going on over here?

The discrepancy comes mostly from the first two terms: the integral from $$2$$ to $$3$$ is about $$2.1$$ while the sum term for $$n=2$$ is about $$4.2$$ Same from $$3$$ to $$4$$ integral is about $$.73$$, the $$n=3$$ sum term is about $$1.1$$ so if you start from $$4$$ on and noting that because the function is decreasing, the integral is always smaller than the sum starting from same bound, you get the integral to be about $$2$$ and the sum to be about $$2.17$$ so the discrepancy gets much smaller.