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What will a good approximation of the following series ?

S=Σ(1/√r) where r varies from 1 to N=10⁶ Specifically, we need to find floor(S)

I tried this by two methods,

1.Using the inequality, S<2√N , which can be proven easily by induction. Using this I got S<2000 but wasn't able to find out floor(S) as S can be 1999 or 1998 or anything below, but close to 2000.

  1. Replacing terms adjacent to perfect squares by perfect squares and adjusting the inequality , i.e replacing √3 by √4 and so on. This was a huge failure as it couldn't work for large values of N.

How would we obtain the right value? Is there a general method for any value of N?

user226375
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    This doesn't help with doing this by hand, but a strong enough calculator can do it easily. – JMoravitz Apr 30 '19 at 17:21
  • See answers of this related question. In particular, we have the asymptotic expansion

    $$\sum_{k=1}^n \frac{1}{\sqrt{k}} \asymp 2\sqrt{n} + \zeta(\frac12) + \frac{1}{2\sqrt{n}} -1/24,{n}^{-3/2}+{\frac {1}{384}},{n}^{-7/2} + \cdots$$ Since $\zeta(\frac12) \sim -1.4603545$, $\left\lfloor \sum_{k=1}^{10^6}\frac{1}{k} \right\rfloor = \left\lfloor 2\sqrt{10^6} - 1.46... \right\rfloor = 1998$

     – achille hui Apr 30 '19 at 17:46

2 Answers2

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We could estimate this sum by using an integral over $f(x)=\frac{1}{\sqrt{x}}$ since $f(x)$ is positive and decreasing over $[1,\infty)$. So

$$\sum_{n=1}^k \frac{1}{\sqrt{n}}\approx\int_1^k \frac{dx}{\sqrt{x}}=2\sqrt{x}|_{x=1}^k=2(\sqrt{k}-1)$$ For $k=10^6$, $$\sum_{n=1}^{10^6}\frac{1}{\sqrt{n}}\approx2(10^3-1)=1998$$

aleden
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    If you compute the sum numerically you get 1998.540 so this is a very good approximation indeed. – cdipaolo Apr 30 '19 at 18:00
  • Shouldn't the correct inequality be, S<∫(1/√x)+1 from 1 to 10⁶ ? While integrating, a lot of extra area will also be added to S so the approximation might work for this case but not for others. Am I right or am I missing something? – user226375 May 01 '19 at 03:10
  • The sum would actually be greater than the integral. In this case we are fortunate that the function is such that the approximation is very close but in general to find how much the two differ you can use the Euler Maclaurin Formula. – aleden May 01 '19 at 03:20
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We can easily prove

$$2(\sqrt{r + 1} - \sqrt{r}) < \frac{1}{\sqrt{r}} < 2(\sqrt{r} - \sqrt{r- 1})$$

So $$2\sum^{10^{6}}_{r=1}\bigg[\sqrt{r+1}-\sqrt{r}\bigg]<1+\sum^{10^{6}}_{r=2}\frac{1}{\sqrt{r}}<1+2\sum^{10^6}_{r=2}\bigg[\sqrt{r}-\sqrt{r-1}\bigg]$$

So $$2\bigg(1000.01-1\bigg)<\sum^{10^6}_{r=1}\frac{1}{\sqrt{r}}<2(10^3-1)+1$$

So we have $$\Bigg\lfloor \sum^{10^6}_{r=1}\frac{1}{\sqrt{r}}\Bigg\rfloor =1998$$

$\text{Added:}$ For $r>1$

$$\begin{gathered} \frac{1}{{\sqrt r }} = \frac{2}{{\sqrt r + \sqrt r }} < \frac{2}{{\sqrt r + \sqrt {r - 1} }} = 2(\sqrt r - \sqrt {r - 1} ) \hfill \\ \hfill \\ \hfill \\ \frac{1}{{\sqrt r }} = \frac{2}{{\sqrt r + \sqrt r }} > \frac{2}{{\sqrt {r + 1} + \sqrt r }} = 2(\sqrt {r + 1} - \sqrt r ) \hfill \\ \end{gathered}$$

DXT
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  • In second step, while changing the limits of Σ , shouldn't it be 2+2Σ(√r-√(r-1)) instead of 1+2Σ(√r-√r-1) ? – user226375 May 01 '19 at 03:34