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Value of $$\bigg\lfloor 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{11}\bigg\rfloor $$

Where $\lfloor x \rfloor$ Represent floor function of $x$.

$\bf{My\; Try::}$ Plotting $\displaystyle y = f(x) = \frac{1}{x}$ in Coordinate axis, We get

$$\int_1^{12}\frac{1}{x}dx<1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots + \frac{1}{11} <1+\int_{1}^{11}\frac{1}{x}dx$$

So we get

\begin{align} \ln(12) & =2.303\log_{10}(12)<1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{11} \\[10pt] & <1+\ln(11) =1+2.303\times \log_{10}(12) \end{align}

How can i solve after that, Help Required, Thanks

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    $\begingroup$ You can just compute the quantity directly, since $1 + 1/2 + ... + 1/11 = 83711 / 27720 \approx 3.02$.... $\endgroup$ – user296602 Jun 6 '16 at 4:20
  • $\begingroup$ I have added like $1+0.5+.33+.125+.......,$ But getting $<3$. $\endgroup$ – juantheron Jun 6 '16 at 4:22
  • $\begingroup$ Then it seems your computation is incorrect somewhere. $\endgroup$ – user296602 Jun 6 '16 at 4:23
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    $\begingroup$ Your approximations by integrals may give away too much. You can do better with $1+\frac{1}{2}+\text{approximations}$. Maybe overkill for $11$. $\endgroup$ – André Nicolas Jun 6 '16 at 4:26
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    $\begingroup$ The integral approximation will not be good enough here as the error between $\int_{k}^{k+1}\frac{{\rm d}x}{x}$ and $\frac{1}{k}$ is $\sim \frac{1}{2k^2}$ which is quite large for all the terms. You need the total error to be smaller than $0.02$ to get the result. Just sum it! More fancy tecniques are going to take much more time anyway. $\endgroup$ – Winther Jun 6 '16 at 4:33
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The exact answer is $$H_{11} = \frac{83711}{27720} > 3$$ so the answer is $3$

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