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Pick the correct option regarding $Q$.

$Q=\frac{1}{100}+\frac{1}{101}+\cdots+\frac{1}{1000}$ Pick one option:

  1. $Q>1\qquad$ 2. $Q\leq \frac{1}{3}\qquad$ 3. $\frac{1}{3} 4. $\frac{2}{3}

My approach :

$Q = \frac{1}{100}+\frac{1}{101}+\cdots+\frac{1}{1000} > \frac{1}{1000}+\frac{1}{1000}+\cdots+\frac{1}{1000} = \frac{901}{1000}$ $\implies \frac{1}{100}+\frac{1}{101}+\cdots+\frac{1}{1000} > \frac{9}{10}$

So, $Q > 1$. Option 1 is correct.

Now, my question is: can this be proved with some other approach?

  • 0
    If you can prove this inequality, $H_n\geq \log(n)\geq H_{n+1}$...2014-04-10

2 Answers 2

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Divide the sum into two parts:

$\frac1{100}+\ldots+\frac{1}{500}+\frac1{501}+\ldots+\frac1{1000}$

The first part, $100$ to $500$ is such that: $\frac1{100}+\ldots+\frac{1}{500}>\frac1{500}+\ldots+\frac{1}{500}=\frac{401}{500}$

The second part, $501$ to $1000$ is such that: $\frac1{501}+\ldots+\frac{1}{1000}>\frac1{1000}+\ldots+\frac{1}{1000}=\frac{500}{1000}$

Therefore the sum of $100$ to $1000$ is larger than $\frac{401}{500}+\frac{500}{1000}$, which itself is larger than $1$.

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Well, let's proceed with the same sort of approach you tried, but a bit finer.

There are $100$ terms greater than $1/200$, another $100$ terms greater than $1/300$, and so on, up to $100$ terms greater than $1/1000$. But this gives that the sum is greater than $1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10$, which is about $1.9$, but more importantly is very easily calculatable to be greater than $1$.

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    Thank you sir, now the concept is crystal clear to me.2012-06-27