Confusion regarding the usage of integration as a limit of a sum

Question

Here is a question I stumbled upon and have solved without the use of integration: What is the greatest integer not exceeding $\sum_{n=1}^{1599} \frac {1}{\sqrt n}$? Below, I've linked another similar type of question and I'm referring to the answer given by user @Hypergeometricx. What is the integer part of $\sum_{i=2}^{9999} \frac {1}{\sqrt i}?$

When I try to convert the summation $\sum_{i=2}^{9999} \frac {1}{\sqrt i}$ into an integral (let me first say $n=9999$), I can first rewrite it as

$\sum_{i=2}^{n}\frac{\frac{1}{\sqrt n} }{\frac {\sqrt i}{\sqrt n}}$ and here I let $\frac{1}{\sqrt n}$ be the infinitesimal element $dx$.

Hence my integral should be $\int_0^{1}\frac 1{\sqrt x}dx\quad$ ; lower limit is approximately $0$ and upper limit is $1$ since the summation is till $i=n$.

However in the answer, the integral has been taken as $\int_2^{10000}\frac 1{\sqrt x}dx\quad $

This is where my confusion lies and I can't really seem to understand why it works. I am not looking for a solution to the problem linked, rather an explanation for this particular method of solving.

Answer 1

I don't think the referenced solution is turning the sum algebraically into a Riemann sum. They're instead using integral bounds, taking advantage of the fact that the function is monotonic. The idea is, the continuous $1/\sqrt{x}$ lies below a left-handed sum, but above a right-handed sum. The technique for infinite series is explained on Khan Academy, but a small modification will get you the upper and lower bounds for finite sums. Let me know if you'd like me to explain further.

Finally, to address your solution, if $1/\sqrt{n}$ is $dx$, then $x=\frac{i}{\sqrt{n}}$ (you have an extra factor of $n^{1/4}$) and $x$ will range from $0$ to $\sqrt{n}$, so those would be the bounds of the integral. It's much more convenient to use $1/n$ as $dx$, then $x=\frac{i}{n}$ and your integral is then just missing a factor of $\sqrt{n}$ and the bounds are correct ($0$ to $1$). Feel free to comment if something is confusing.