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

Question

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?

Answer 1

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.