Give it a try

Here, we try to find the upper & lower bounds of the sum. For clarity, refer to the graph given below.

• The sum, S will be the area under $\frac{1}{\lfloor{x}\rfloor}$ (blue).
• Upper bound of S will be the area under upper curve, $\frac{1}{x-1}$ (green)
• Lower bound of S will be the area under the lower curve, $\frac{1}{x}$ (red)

$\int_{10}^{101}\frac{1}{x-1}>S>\int_{10}^{101}\frac{1}{x}$ $log\frac{100}{9}>S>log\frac{101}{10}$ $2.41>S>2.31$

The only choice in that bound is 2.31 .

We may use the following approximation,

$S = \frac{1}{d} \times ln\frac{2a+(2n-1)d}{2a-d}$

Where,

The series may be as,

$\frac{1}{a},\frac{1}{a+d},...,\frac{1}{a+(n-1)d}$

However, by this method the answer turns out to be $2.35$ ,and so I took the best shot choosing the nearest option.😀

We use the well-known approximation $H_n = \frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{n} \approx \gamma + \ln n$ , where $\gamma$ is the Euler-Mascheroni constant. We thus have

$S = H_{100}-H_9 \approx \gamma + \ln 100 - \gamma - \ln 9 = \ln \frac{100}{9} \approx 2.408$

Thus, we pick the closest answer, which happens to be $\boxed{2.31}$ .

This isn't a new question. Despite a more exact value of 2.35840926367136, 2.30258509299405 is taken as an approximation.

$\displaystyle \int_{10}^{100}\frac{1}{x} d x$ = Ln ( $\frac{100}{10})$ = Ln 10

Since there is only one nearest option, Ln 10 above is sufficient to determine the choice of 2.31.

Answer: 2.31

I did the same like harish sasikumar but I want to make an addon ...is it correct that we are doing this approximation by taking 1 unit width rectangles and so this width is making a larger bracket of limit...also is it possible through algebra that we get this exact answer from the first 2 decimal places??