Floor sum of reciprocal of square root of first $50$ numbers

Question

Find Sum of $$\bigg\lfloor 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\cdots+\frac{1}{\sqrt{50}}\bigg\rfloor$$

$\bf{My\; Try::}$ Let $\displaystyle y=f(x) = \frac{1}{\sqrt{x}}\;,$ Then draw that graph in coordinate axis, We get

$$\displaystyle \int_{1}^{51}\frac{1}{\sqrt{x}}dx<\sum^{50}_{k=1}\frac{1}{\sqrt{k}}<1+\int_{1}^{50}\frac{1}{\sqrt{x}}dx$$

So we get $$\displaystyle 2\left(\sqrt{51}-1\right)<\sum^{50}_{k=1}\frac{1}{\sqrt{k}}<1+2(\sqrt{50}-1)$$

So we get $$12.28<\sum^{50}_{k=1}\frac{1}{\sqrt{k}}<13.14$$

But answer given is $12,$ How cai i solve avobe question, Help Required, Thanks

Answer 1

By the same method, you find for $m>1$ $$ 2(\sqrt{51}-\sqrt m)<\sum_{k=m}^{50}\frac1{\sqrt k}<2(\sqrt{50}-\sqrt{m-1})$$ With $m=5$ (actually, $m=4$ would be enough, but is "harder to compute" as it involves more numerical square root computations), this gives us $$\begin{align}\sum_{k=1}^{50}\frac1{\sqrt k}&<1+\frac1{\sqrt 2}+\frac1{\sqrt 3}+\frac1{\sqrt 4}+2(\sqrt{50}-\sqrt 4)\\ &<1+0.71+0.58+0.5+14.15-4 \\ &=12.94\end{align}$$

Answer 2

$$ \sum_{k=1}^{50} \frac 1 {\sqrt{k}} = \int_1^{51} \frac 1 {\sqrt{x}} \, dx + \text{a little bit more} = 2(\sqrt{51} - 1) + \text{a little bit more}. $$ We just need to show that that "little bit" is small enough. A carefully drawn graph show show you why $$ \text{that little bit} < \left( 1 - \frac 1 {\sqrt 2} \right) + \left( \frac 1 {\sqrt 2} - \frac 1 {\sqrt 3} \right) + \left( \frac 1 {\sqrt 3} - \frac 1 {\sqrt 4} \right) + \cdots + \left( \frac 1 {\sqrt{50}} - \frac 1 {\sqrt{51}} \right) $$ and all of the terms cancel out except the first and the last, so $$ \text{that little bit} < 1 - \frac 1 {\sqrt{51}}. $$ That doesn't quite do it, so we refine the technique: $$ \frac 1 {\sqrt 1} = \int_1^2 \frac{dx}{\sqrt x} + \text{a little bit} = 2(\sqrt 2 - 1) + \text{a little bit}. $$ That little bit can be found numerically and it is less than $60\%$ of $1 - \dfrac 1 {\sqrt 2}$.

In all the latter terms, the $60\%$ would be replaced by something even smaller than $60\%$ (but always bigger than $50\%$ for reasons that should be obvious from looking at the graph).

Thus $$ \text{little bit} < 0.6\times \left( 1 - \frac 1 {\sqrt 2} \right) + \left( \frac 1 {\sqrt 2} - \frac 1 {\sqrt 3} \right) + \left( \frac 1 {\sqrt 3} - \frac 1 {\sqrt 4} \right) + \cdots + \left( \frac 1 {\sqrt{50}} - \frac 1 {\sqrt{51}} \right) $$ and that does it.

Answer 3

It's lots easier if you use centered integrals. A little algebra shows that $$\begin{align}\int_{k-\frac12}^{k+\frac12}\frac{dx}{\sqrt x}-\frac1{\sqrt k}&=\frac1{2\sqrt k\left(\sqrt{k+\frac12}+\sqrt{k-\frac12}\right)^2\left(\sqrt{k}+\sqrt{k-\frac12}\right)\left(\sqrt{k+\frac12}+\sqrt{k}\right)}\\ &\le\frac1{32\left(k-\frac12\right)^{5/2}}\end{align}$$ So evaluating the first error exactly and bounding the rest by the number of terms times the largest error, $$0\le\int_{\frac12}^{\frac{101}2}\frac{dx}{\sqrt x}-\sum_{k=1}^{50}\frac1{\sqrt k}\le0.03527618+49\cdot0.005681136=0.313651854$$ $$\int_{k-\frac12}^{k+\frac12}\frac{dx}{\sqrt x}=2\sqrt{\frac{101}2}-2\sqrt{\frac12}=12.798457$$ So $$12.484805\le\sum_{k=1}^{50}\frac1{\sqrt k}\le12.798457$$ So the sum of centered integrals was accurate enough to determine the floor of the sum. Of course, we had to accurately bound the error...

EDIT: There is a much less strenuous way to develop a useful error bound. Start with the Taylor series $$2\sqrt{k+\frac12}=2\sqrt k+2\frac{\left(\frac12\right)}{1!}\frac{\left(\frac12\right)}{k^{1/2}}+2\frac{\left(\frac12\right)\left(-\frac12\right)}{2!}\frac{\left(\frac12\right)^2}{k^{3/2}}+2\frac{\left(\frac12\right)\left(-\frac12\right)\left(-\frac32\right)}{3!}\frac{\left(\frac12\right)^3}{\xi_{2k}^{5/2}}$$ $$2\sqrt{k-\frac12}=2\sqrt k+2\frac{\left(\frac12\right)}{1!}\frac{\left(-\frac12\right)}{k^{1/2}}+2\frac{\left(\frac12\right)\left(-\frac12\right)}{2!}\frac{\left(-\frac12\right)^2}{k^{3/2}}+2\frac{\left(\frac12\right)\left(-\frac12\right)\left(-\frac32\right)}{3!}\frac{\left(-\frac12\right)^3}{\xi_{2k-1}^{5/2}}$$ With $\xi_{2k}>k+\frac12$ and $\xi_{2k-1}>k-\frac12$. Simplifying and subtracting, $$2\sqrt{k+\frac12}-2\sqrt{k-\frac12}=\frac1{\sqrt k}+\frac1{8\sqrt2}\left(\frac1{(2\xi_{2k-1})^{5/2}}+\frac1{(2\xi_{2k})^{5/2}}\right)$$ So that way our error estimate reads $$0\le\int_{\frac12}^{\frac{101}2}\frac{dx}{\sqrt x}-\sum_{k=1}^{50}\frac1{\sqrt k}=\frac1{8\sqrt2}\sum_{k=1}^{100}\frac1{(2\xi_k)^{5/2}}<\frac1{8\sqrt2}\sum_{k=1}^{100}\frac1{k^{5/2}}<\frac1{8\sqrt2}\sum_{k=1}^{\infty}\frac1{k^2}=\frac{\pi^2}{48\sqrt2}$$ So now we have $$12.798457-\frac{\pi^2}{48\sqrt2}=12.653064<\sum_{k=1}^{50}\frac1{\sqrt k}<12.798457$$ So this way we developed an analytical error bound of sufficient accuracy without any real numerical work, just a simple formula. This, along with evaluation of the primitive at the endpoints, was sufficient to establish that $$\left\lfloor\sum_{k=1}^{50}\frac1{\sqrt k}\right\rfloor=12$$

Answer 4

May be, you could have used $$\sum_{i=1}^n \frac 1 {\sqrt n}=H_n^{\left(\frac{1}{2}\right)}$$ where appear the generalized harmonic numbers and then have used their asymptotic expansions $$H_n^{\left(\frac{1}{2}\right)}=2 \sqrt{n}+\zeta \left(\frac{1}{2}\right)+\frac 1 {2\sqrt{n}}-\frac{1}{24n\sqrt{n}} +O\left(\frac{1}{n^{5/2}}\right)$$

For $n=50$, limiting to $O\left(\frac{1}{n^{1/2}}\right)$, this would give $\approx 12.6818$; limiting to $O\left(\frac{1}{n^{3/2}}\right)$, this would give $\approx 12.7525$; imiting to $O\left(\frac{1}{n^{5/2}}\right)$, this would give $\approx 12.7524$.