Problem: Find the integer part of $$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} +...+\frac{1}{\sqrt{1000000}}$$
How should I go about approaching this question? I have never seen such a question before, where you cannot multiply any conjugate or whatsoever. What is the method to go about approaching such problems?
For an precalculus (telescoping) approach. Notice that $$\frac{1}{\sqrt n}>\frac{2}{\sqrt{n}+\sqrt{n+1}}=2(\sqrt{n+1}-\sqrt{n})$$ From this $$\frac{1}{\sqrt{1}}+\frac1{\sqrt2}+\cdots+\frac1{\sqrt{1000^2}}\geq 2(\sqrt 2-1+\sqrt 3-\sqrt 2+\cdots+\sqrt{1000^2+1}-\sqrt{1000^2})\geq 2\sqrt{1000^2+1}-2>2000-2=1998$$ And also $$\frac{1}{\sqrt n}\leq\frac{2}{\sqrt n+\sqrt{n-1}}=2(\sqrt n-\sqrt{n-1})$$ Leaving the $n=1$ case alone we get $$\frac{1}{\sqrt{1}}+\frac1{\sqrt2}+\cdots+\frac1{\sqrt{1000^2}}<1+2(\sqrt 2-1+\sqrt 3-\sqrt 2+\cdots \sqrt{1000^2}-\sqrt{1000^2-1})=1+2\sqrt{1000^2}-2=1999$$
And it follows that the integer part is $1998$
By integration of
$$\frac1{\sqrt x}\le\frac1{\sqrt{\lfloor x\rfloor}}<\frac1{\sqrt{x-1}}$$
we establish a formula for the tail of the summation,
$$\int_m^{n+1}\frac{dx}{\sqrt x}\le\sum_{k=m}^n\frac1{\sqrt k}<\int_m^{n+1}\frac{dx}{\sqrt{x-1}},$$ or
$$2(\sqrt{n+1}-\sqrt m)\le\sum_{k=m}^n\frac1{\sqrt k}<2(\sqrt n-\sqrt{m-1}).$$
Then for the complete sum
$$\sum_{k=1}^{m-1}\frac1{\sqrt k}+2(\sqrt{n+1}-\sqrt m)\le\sum_{k=1}^n\frac1{\sqrt k}<\sum_{k=1}^{m-1}\frac1{\sqrt k}+2(\sqrt n-\sqrt{m-1}).$$
We can compute the bracketing for increasing $m$, and it turns out that $m=2$ suffices to establish
$$1998.17257288\le S<1999.0$$
Hence, $$\color{green}{1998}.$$
For safety, with $m=3$,
$$1998.24400517\le S<1998.87867966$$
Rationale of the method:
An integral approximates a sum inasmuch as the function value remains sufficiently constant in unit intervals. In the case of the inverse square root, the error per interval is
$$\frac1{\sqrt{n+1}}-\frac1{\sqrt n}=O(n^{-3/2}),$$ which decreases relatively quickly. So there is some hope that by combining a direct summation of the first terms and a bracketing of the tail, we can end up with the two bounds in the same unit interval so that we know the integer part. (The last terms won't be a problem as they correspond to a tiny error.)
We are pretty lucky here as the unit interval is found with very few terms.
Given that the width of the bracketing is
$$2(\sqrt n-\sqrt{n+1}+\sqrt m-\sqrt{m-1}),$$ we can conclude as soon as the partial sum differs from an integer by more than this amount. And the firt parial sums are approximately
$$1,1.71,2.28,2.78,3.23\cdots$$
which leave a good margin.
If you know about generalized harmonic numbers $$S_n=\sum_{k=1}^n \frac 1 {\sqrt k}=H_n^{\left(\frac{1}{2}\right)}$$ Now, using the asymptotics $$S_n=2 \sqrt{n}+\zeta \left(\frac{1}{2}\right)+\frac1 {2\sqrt{{n}}}+O\left(\frac{1}{n^{3/2}}\right)$$ where $\zeta \left(\frac{1}{2}\right)\approx -1.46035$.
SO, if $n=10^6$, the approximation gives for your summation $\approx \zeta \left(\frac{1}{2}\right)+\frac{4000001}{2000}\approx 1998.54$ while the rigorous calculation would give ... the same value !
Edit (five years later)
To answer a question asked in private, considering the more general case of $$S_n=\sum_{k=1}^n \frac 1 {\sqrt[a] k}=H_n^{\left(\frac{1}{a}\right)}$$ using asymptotics $$\color{blue}{n^{\frac{1}{a}} \left(S_n-\zeta \left(\frac{1}{a}\right)\right)=}$$ $$\color{blue}{\frac{a n}{a-1}+\frac{1}{2}-\frac{1}{12 a n}+\frac 1{720}\sum_{k=1}^\infty (-1)^{k+1}\, \frac{\prod_{m=1}^{2k} (a m+1) } {\alpha_k\,(an)^{2k+1} }}$$
where the $\alpha_k$ form the sequence $$\{1,42,1680,66528,1816214400,103783680,14820309504000,\cdots\}$$
Multiplied by $720$, you will find them in sequence $A060055$ in $OEIS$.
