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?


3 Answers 3


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


which leave a good margin.

  • 2
    $\begingroup$ Simple and nice (as usual). Cheers. $\endgroup$ Jan 11, 2018 at 11:36
  • 2
    $\begingroup$ @ClaudeLeibovici: greetings, Claude. My secret (maybe I already said it): I am unable to do complicated things ;-) $\endgroup$
    – user65203
    Jan 11, 2018 at 11:38
  • $\begingroup$ For instance if it were to change to + then -, would the approach be the same? $\endgroup$
    – QuIcKmAtHs
    Jan 11, 2018 at 11:54
  • 1
    $\begingroup$ @XcoderX: then you would take the terms two by two to get rid of the alternation (and get a yet faster convergence). $\endgroup$
    – user65203
    Jan 11, 2018 at 11:55
  • 1
    $\begingroup$ @WillJagy. Thanks for the link. It is very interesting (question and answers). $\endgroup$ Jan 12, 2018 at 4:37

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 $OEIS$.


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