# Which of these statements about $Q=\frac{1}{100}+\frac{1}{101}+\cdots+\frac{1}{1000}$ is correct?

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

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?

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Now, My question is can this be proved with some other approach? –  Bazinga Jun 27 '12 at 21:54
Simply knowing that $Q>9/10$ is insufficient to conclude that $Q>1$. You've not ruled out option $4$, yet. –  Cameron Buie Jun 27 '12 at 21:56
Right, so everything except 4 and 1 have been ruled out. –  Eric Stucky Jun 27 '12 at 21:59
Sorry everyone. Can you guide me further? –  Bazinga Jun 27 '12 at 22:00
To complete the proof, you could break the series at 200: $1/100 + \dots + 1/200 > 101/200 > 1/ 2$, and then $1/201+\dots + 1/000 > 800/1000 = 4/5$, and those add up to more than 1. I'm not aware of a fundamentally different approach that is accessible at a precalculus level. –  Eric Stucky Jun 27 '12 at 22:04

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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Sir, I got your last line but please elaborate "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," –  Bazinga Jun 27 '12 at 22:15
For example, $1/100 + 1/101 + \dots + 1/199 > 100/200 = 1/2$, and repeating we get that the whole sum is greater than $1/2 + 1/3 + \dots + 1/10$ –  mixedmath Jun 27 '12 at 22:20
Thank you sir, now the concept is crystal clear to me. –  Bazinga Jun 27 '12 at 22:37